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1^ 



Carnegie XTecbnical Scbools Zcxt Boofes 



MATHEMATICS 



FOR 



ENGINEERING STUDENTS 



BY 

PROF. S. S. KELLER 

CARNEGIE TECHNICAL SCHOOLS 



ALGEBRA AND TRIGONOMETRY 




NEW YORK: 

D. VAN NOSTRAND COMPANY 

23 MURRAY AND 27 WARREN STS. 
I907 






ov* 



LIBRARY of CONGRESS 
Two Cooles Received 

APh 19 1907 

«ooyn«ht Entry 

, / % ' f ° 7. 

CLASS A XXc, No. 
COPY B. 



Copyright, 1907, by 
D. VAN NOSTRAND COMPANY. 



Stanbope press 

. GILSON COMPAN 
BOSTON, U. S. A. 



PREFACE. 

Although this book has been designed to meet the 
specific needs of the Carnegie Technical Schools, the grow- 
ing demand in the field of technical education for a form of 
mathematical instruction that will eliminate the purely 
speculative and concentrate the more utilitarian features 
of mathematical science, leads the author to believe that 
such a work as this will not be entirely inept outside of 
the Carnegie Technical Schools. 

It is believed that the intellectual stimulus and dis- 
cipline that is usually attributed to mathematical studies 
can be as readily conveyed by those things that are at the 
same time practically useful, as by those that are merely 
speculative ; perhaps much more readily. 

The child can be taught to read as successfully by 
giving it exercises that contain useful information, as by 
requiring it to drone over masterpieces of prose and verse 
that leave no valuable residue whatever in the childish 
mind. 

The subjects discussed and the problems illustrative of 
them have been selected after a careful gleaning of the 
author's experience with students of varying tastes and 
mentality, with the end in view of making the subjects 
vital and pertinent to the special training they are seeking, 
and at the same time of developing their powers of inde- 
pendent and accurate thinking. 

The vital thing in the art of instruction, in the author's 
opinion, is to retain for the subject under investigation, 



iv Preface. 

not only the students interest but his respect and confi- 
dence. To that end he must feel that he is not simply- 
grinding thin air to make, too often, an intellectual fog. 

An effort has been made to avoid the extreme of 
pruning too closely, and, in consequence, everything that 
in the author's judgment can have even a remote bearing 
upon a student's usefulness in technical pursuits has 
been inserted more or less briefly. 

The writer wishes to acknowledge his indebtedness to 
Mr. W. A. Bassett and Mr. Lightcap, instructors in the 
Carnegie Technical Schools, and especially to Professor 
Walter F. Knox, for valuable assistance and suggestions. 



ALGEBRA. 



ALGEBRA. 

CHAPTER I. 

ALGEBRA AND ARITHMETIC. 

Article i. Algebra is merely an extension of the field 
of Arithmetic, the primary difference being the use of 
letters as symbols, in addition to that of the Arabic char- 
acters (i, 2, 3, 4, etc.) and the general employment of 
equations. 

Fundamental Operations. 

Art. 2. The arithmetical operations called addition, 
subtraction, multiplication, and division, have the same 
meaning in algebra. For- instance, in arithmetic we 
would understand by the symbols 5 + 4, that 5 and 4 are 
added ; so in algebra by the symbols a + b, we would under- 
stand that a quantity represented by a was added to 
another quantity represented by b. Also a - b would 
mean, as in arithmetic, that b was to be subtracted 
from a. 

Negative Numbers. 

Art. 3. In arithmetic it is always necessary in sub- 
traction that the quantity subtracted (called the sub- 
trahend) be less than the quantity from which it is taken 
(called the minuend). 

The use of letters, especially when unknown quantities 
are thus represented, makes desirable an extension of the 
operation of subtraction, making it always possible. 

The desirability of representing in symbols such 



2 Algebra. 

balancing conditions as debit and credit, profit and loss, 
above zero and below zero, etc., has suggested the idea of 
negative numbers. 

Illustration. A business man whose assets are $9000 
loses $10,000; he not only has nothing left, but is $1000 
in debt. 

To express this condition the idea of negative numbers 
must be introduced, and it is said that he is worth 
— $1000. 

Again, A travels from C to D 5 miles. At D, his 
distance from C is 5 miles then. He returns from D to C, 
and each mile he travels toward C reduces his distance 
from C one mile, and he is successively 4, 3, 2, 1, and o 
miles from C ; the last symbol (zero) indicating that he 
has arrived at C. Suppose he is carried through C to E 
2 miles, then we may say he is — 2 miles from C. 

The idea of positive and negative quality may thus be 
represented by opposite directions, from a fixed point. 
Let this idea be applied to the series of numbers, o, 1, 2, 3, 
4, 5, 6, 7, 8, 9, etc. It is plain that this series can be 
extended to the right indefinitely by adding 1 to each 
successive number, but if, starting from any number in 
this series, we return toward the left, by subtracting 1 
each time we cannot extend this process indefinitely, 
because the series of arithmetical numbers, here repres- 
ented, ends on the left with what we call zero. 

Now suppose we agree to extend the series to the left 
of zero, still subtracting 1 each time, we must have some 
new designation for the resulting numbers, and we agree 
to call them negative numbers and to represent the series 
thus : 

As -8,-7,-6,-5,-4,-3,-2,-1 

o, 1, 2, 3, 4, 5, 6, 7, 8, etc. 

We can now express the condition of the business man 
referred to above. He has lost not only his $ 9000, which 



Algebra. 3 

would stand in the series to the right of o, by successive 
subtractions, but these subtractions still continued until 
the number representing the final condition stood far 
down in the series to the left, that is, was— iooo. 

From the method of obtaining this idea of negative 
numbers it is plain that they are all less than zero. 

A practical illustration in point is the thermometer 
scale ; below zero readings being represented as negative 
numbers; thus, — 5 , — io°, etc., mean 5 degrees and 
10 degrees below zero, respectively, etc. 



SYMBOLS. 

Art. 4. The signs of addition, subtraction, multiplica- 
tion, and division (+ - X -5-) are the same as in arith- 
metic, and have the same general significance. 

Art. 5. The greater necessity for indicating opera- 
tions that cannot be completely performed in algebra 
makes frequent use of the parenthesis or bracket neces- 
sary [(.••)[•••]{•••}]• I 1 * s often desirable to treat 
a polynomial, for instance, as if it were a monomial, and 
one of the important functions of the parenthesis is to 
bind together several terms where they are to be treated 
like a single term. For instance, if we wanted to show 
that the expression x 3 y' 6 + 3.V 2 y 2 + Gxy + 7 was like the 
expression a 3 + 3a 2 4- 6a + 7 in form, it can be written 
thus, (xy) 3 + $(xy) 2 + 6(xy) +7. Or, that x + y + z + 
ax + ay + az was like m + am, the first expression can 
be written (x + y + z) + a (x + y + 2). 

Art. 6. The expression 4a 2 -(5a + 2b — c) means that 
the expressions 5a, 2 b, and-c are all to be subtracted 
from 4a 2 . The subtraction is merely deferred by the use 
of the parenthesis until it is desirable to actually perform 
it. As long as the parenthesis embraces an expression, 
its parts are inseparable and must be taken together. 



4 Algebra. 

ADDITION. 

Definitions. 

Art. 7. Since the use of letters makes it necessary 
to merely indicate the operations of addition, subtraction, 
ihultiplication, etc., it is necessary to add to the language 
of arithmetic certain names for these new relations. For 
instance, in the expression $ax, 5 is called the coefficient 
of ax; the expression might have been written a$x, then 
a would have been called the coefficient of the expres- 
sion ; likewise in x$a, x is the coefficient.* 

It is customary, however, when a number is present in 
an expression to regard it as the coefficient. 

From this description, formulate a definition for coeffi- 
cient. 

Art. 8. A single expression, involving letters or num- 
bers in any amount, whose parts are not connected by 
plus or minus signs is called a monomial. 

Art. 9. Two or more monomials joined by plus or 
minus signs form a polynomial. Each monomial is called 
a term of the polynomial. 

A polynomial of two terms is called a binomial ; one 
of three terms, a trinomial, etc. 

Art. 10. Monomials are said to be like when they 
differ, if at all, only in their numerical coefficients. For 
example, 5a 2 xy, 6a 2 xy, 10a 2 xy, etc., are alike. 

Give an example of a monomial, of a binomial, of a 
trinomial. 

Addition of Monomials. 

Art. 11. When monomials are like, each consists of 
the same letters, affected in exactly the same way ; the 
numerical coefficients simply indicate how many of these 

* It is sometimes necessary to extend this idea of a coefficient 
to include any group of letters or numbers or both, in an expression. 
For instance, 5a might be called the coefficient of x, or in ^a 7 dc, 3# 2 # 
might be called the coefficient of c, etc. 



Algebra. 5 

same groups of letters each monomial contains. Plainly- 
then to add monomials ; if they are like, add the coefficie?its 
with their proper signs and attach the sum as coefficient to the 
common literal part ; if unlike , connect them by their proper 
signs. 

It is clearly impossible to collect unlike monomials into 
one expression, as it would be arithmetically impossible to 
add pounds, inches, and pints. Unlike monomials are es- 
sentially different things, and their addition or subtraction 
can only be indicated, not actually performed. 

Addition of Polynomials. 

Art. 12. As polynomials are made up of monomials, 
their addition resolves itself into an addition of the mo- 
nomials that compose them. To facilitate this, it is desir- 
able to arrange the like monomials, which occur in the 
polynomials to be added, in columns. Then add the like 
monomials according to their rule of addition, and to these 
sums join all the unlike mono?nials occurring in all the poly- 
nomials, with their p?'Opcr signs. 

For example : 

Add 3a 2 x — 2aby + jxy 2 — 6a 2 and a 2 x + $xy 2 + 3b 2 . 

Arrange, 3a 2 x - 2aby + jxy 2 - 6a 2 

a 2 x + $xy 2 + 3b 2 



\a 2 x — 2aby + i2xy 2 — 6a 2 + 3b 2 

Subtraction of Monomials. 

Art. 13. From what has been said under addition it 
follows ; to subtract like monomials : Subtract algebraically 
their numerical coefficients and attach the common literal part 
to the remainder. 

If the monomials are unlike : Change the sign of the sub- 
trahend a fid join it to the minuend. 



Algebra. 



To Subtract Polynomials. 



Art. 14. A rrange like terms under each other and sub- 
tract according to rule jor monomials. 

For example: From 2x + 17X 3 — 20 take — $x 2 — 15 
+ 39*- 3 

Arrange, 2x + ijx 3 —20 

Subtract, - 3# 2 + 39X 3 - 15 

3# 2 + 2x — 2 2x 3 — 5 

ADDITION OF NEGATIVE QUANTITIES. 

Art. 15. The addition of negative quantities adds a 
new concept to the theory of addition. 

If we are required to find the value of the expression 
7 + (—2) we may have recourse to the series of positive 
and negative numbers previously referred to ; thus : 

+1+1+1+1+1+1+1 + 1 +1+1+ 1+1+1 +1+1+1+1+1 

— O— S— *— G -3— 4— 3— i— 1 ± ©+1+3+3+4+5+0+1+8, 
—1—1—1—1—1—1—1—1—1— 1 —1-1—1—1—1—1—1—1 



It will be observed that the numbers increase by unity 
to the right, as indicated by the upper row of i's and 
decrease to the left, as indicated by the lower row. 

Now, suppose that it was required to find 7 +(—2). 
This indicates that we must count two to the left by the 
lower row from + 7, which gives +5. If it was required 
to find 7 + ( - 9) we must count 9 units to the left, which 
brings us to —2, etc., hence we say 7+ ( — 2) =5, and 
7+( — 9)=— 2; but 7+ (-2) = 5 is the same as 
7 — 2 = 5, which is simple arithmetical subtraction, but for 
7+( — 9) = — 2 we have no arithmetical equivalent. From 
an observation of the example and others like it, using the 
series, we are enabled to evolve this rule : 



Algebra. 7 

To add negative numbers subtract the less number from 
the greater, arithmetically, and prefix the sign of the greater 
number. 

Examples. 5-8= - 3, 

that is, 5 is subtracted from 8 and the sign of 8 is pre- 
fixed. 

+ 9-2= + 7, etc. 

This rule may be extended to a series of quantities by 
combining the positive quantities into one group and the 
negative into another, thus : 

7-5+11-3-10+13-1 = (7+ 11 + 13) + (-5-3-10 
-1) =31 + (-19) = +31-19= +12, etc. 

Subtraction of Negative Numbers. 

Art. 16. In the expression a- b = c, b is evidently the 
number which added to c will give a, and in general we 
understand that the subtrahend added to the remainder 
must always give the minuend. Applying this principle to 
negative numbers 7 -(-2) =9, for -2 is the quantity 
added to 9 gives 7, according to the rule for addition 
already stated. Again, — 5 — ( — 3) = — 2, for — 3 added 
to - 2 gives -5. But 7 + 2 also equals 9 and -5 + 3 
also equals — 2 ; hence we may express the rule : 

To subtract a negative (or positive) quantity change its 
sign and add. 

EXERCISE I. 

Addition. 

Find sum in each following example: 

1. 2a - 3X 2 , $x 2 - 7a, - 3a + x 2 , and a - 3X 2 . 

2. m 2 -n 2 + 3tn 2 n- 5mn 2 , 3m 2 - 4m 2 n+ 3W 3 - stnn 2 , m 3 + 

n 3 +T,m 2 n, 2m 3 — 411 3 — $mn 2 , 6m 2 n+ lomn 2 , and — 
6m 3 - jmhi + %mn 2 + 2ti 3 . 



8 Algebra. 

3. 2>y^~ 4-y^+zy 2 , $y 2 a—mn + y\, 4y 2 — $y 3 , and sy$ — 4 

+ yh 

4. 6-7 -66 2 + 14& 3 , 6+ i6b*-<)b-2b 2 , and 4ft 2 - 9ft 3 + 

130 + 10. 

5 - §# — £? + T42 and - \x + ly - f z. 

6. 13 (a+2b)-i$ (2b + c), 6 (2b + c)-j (c+a), and 

2 (c + a) -5 (a + 2 6). 

7. 2^ 2 z - $yz 2 + $y 3 , %yz 2 - 42 s + jy 2 z, yy 3 - 52 s - %yz 2 , and 

-62 s — $y 2 z+ 7y 3 . 

8. 8a 3 — 1 1a — ja 2 , 2a - 6a 2 + 10, — 5 + 4# 3 + 9 a, and 13a 2 

-5- 12a. 

9. 5W 2 — 13m- 4+ 5m 3 , 7m 3 — 7wz 2 + 2m— 14, 6w 3 + 8 — 

iom — Sm 2 , and 15 - i6m 2 + 15m -m 3 . 

10. 3*fyz - x^ 2 2 + 72 2 , 92^ - x 2 ^2 + Jxy 2 z, 2z 2 y — zz 2 + 5x^2, 

and 2 2 -4x^2. 

11. tx 2 -fx-f^ + i hx 2 -\ + lx + %y, and ±y-s x + \% 

-tW- 

12. .2a); 2 -.3a 2 + .o6y 3 m + 7i 3 , ^ay 2 + .2$y 3 m- .\a 2 - -35« 3 , 

.3a)/ 2 + .05W 3 — a 2 — .o4;y 3 ra, and — .6n 3 + .3a 2 
— .15 y 3 m —ay 2 . 

13. Simplify, 8mx — $x 2 + 3m 2 + 2x 2 — &m 2 + 13m 2 — i8mx 

+ 6x 2 — 9m 2 . 

14. Add, 2n 3 + fan 2 — ^a 2 n + 3a 3 , 8a 2 n — 15 aw 2 — 5a 3 — 

iow 3 , ^n 3 -6an 2 , and - an 2 + n 3 - 4a 3 . 

EXERCISE II. 

Subtraction. 
Subtract : 

1. sa 2 y — 2 bx 2 - x 3 from 2a 2 ^ + 3x 3 — 4&X 2 . 

2. x 3 - 3x 2 >> + 2>ocy 2 - y 3 from 2x 3 — 2x 2 y + \xy*. 

3. m 3 + 1 from 3m 2 + 2m — 6. 

4. 3X 3 - \x 2 + 2x from 2x 3 — x 2 . 

5. a§ — 3a*^+M from 3a§+6a*^~55i 



Algebra. 9 

6. 3 (m + n) + 2 (x + y) from 5 (m + n)-y (x + y). 

7. ax+by + c from mx + ny — d. 

8. What must be added to 2x 2 - $xy-y 2 to make x 2 + 

239/ + ;y 2 ? 

9. Perform the indicated operations; 43; — $y — [2x— 5 

-(3^-^-2)]. 

10. 3a 2 -[7a- (a 2 - 2a + 9)]. 

11. Subtract 7x3/ + 3^ — 4yx + 2y from 9x3^ — 5^ + |y and 

add to the remainder xy — x + \y. 

12. Simplify, $x + 2y — (9V — jx) — ( — x — y). 

Multiplication. 

Art. 17. In the series of numbers it may be conceived 
that a number may be revolved from the positive side of 
zero to the negative side, or vice versa. Thus in the 
series, 

-9-8-7-6-5-4-3-2-1, o, 
+ 1 + 2 + 3 + 4 + 5 + 6+7 + 8+ + 

+ 6 may be carried over as if revolved on an arm pivoted 
at o, to — 6 on the other side, but that is equivalent to 
multiplying 6 by - 1 ; or + 3 may be revolved to - 6, 
which corresponds to multiplying 3 by — 2, etc. Hence, 
multiplying by a negative quantity carries the number 
over to the other side of zero. 

Suppose — 6 is multiplied by - 1, then by this rule the 
product would be + 6, for — 6 would be revolved to + 6 
on the other side, or — 3 multiplied by — 2 would give us 
+ 6 ; hence, two negative quantities multiplied together pro- 
duce a positive quantity. 

A general rule may be stated thus : 

To multiply any two quantities, multiply the quantities 
independent of their signs, then prefix a sign determined 
by the rule that like signs give + and unlike signs — . 



10 Algebra. 



Division. 

Art. i 8. Division being the reverse of multiplication, 
the rules governing its application are derived from the 
rules governing multiplication. If it is required to find 
i6~-2, it is necessary only to find the number which 
multiplied by 2, gives 16, which is 8. Likewise, 28 h — 4 
= — 7, because — 7 is the number by which — 4 must be 
multiplied to equal 28. 

Again, — 32 -. — 4 = 8, for like reason ; also — ab + b = 
— a, or — abc -. — bc= + a, etc. Hence, to divide, find the 
quotient of the quantities independent of sign, and prefix 
the sign determined by the rule that like signs give + 

AND UNLIKE SIGNS, — . 

Exponents. 

Art. 19. When a quantity is multiplied by itself two 
or more times, as a X a X a X a, the result is represented by 
an abbreviated notation, as a 4 in the above instance. The 

4 is called an exponent and indicates the number of times 
the quantity (as a) called the base, is repeated, thus: (ab) 5 
means that (ab) has been multiplied by itself 5 times, 

5 being called the exponent, and ab, the base. In 
general such an expression is called a power of the quan- 
tity, and the power is said to be an even power when the 
exponent is an even number, and odd, when it is an odd 
number. 

Laws of Exponents. 

Art. 20. If it is required to find the product a 4 X a 5 , 
it is understood that we have a product of "a" multiplied 
by itself 4 times, multiplied by another product of a multi- 
plied by itself 5 times, which is plainly the same thing as 
a single product of a multiplied by itself 9 times. But 
9 = 4+55 that is, the exponents are added. 



Algebra, 1 1 

Hence, in multiplying expressions containing the 

SAME QUANTITY AFFECTED BY DIFFERENT EXPONENTS, ADD 
THE EXPONENTS OF THAT QUANTITY IN THE SEVERAL EX- 
PRESSIONS TO OBTAIN ITS EXPONENT IN THE PRODUCT. 

For example, 

a 2 b 3 c X ab 2 c 4 = a 2+1 b 3 + 2 c l + 4 = a 3 b 5 c 5 . 

Let it be required to find a 8 -?- a 3 . This expression 
means that a quantity is to be found that multiplied by 
a 3 = a 8 . By the law of exponents for multiplication, 3 + 
some number =8, from which that number required 
= 8-3 = 5. B ut 8 - 3 = the difference of the exponents 
involved, hence, to divide two expressions containing the 
same quantities, subtract the exponents oj the quantities in 
the divisor from the exponents of the same quantities in the 
dividend, and the remainders will be the exponents 0} those 
quantities in the quotient. 

Thus, a 4 b 3 c 2 d 5 -h a 2 bd 3 = a 4 - 2 b 3 - l c 2 d"- 3 = a 2 b 2 c 2 d. 

Again, a n b m c 3 d 4 -*■ a 2 b r c v d 2 = a n - 2 b m - x c 3 ~ v d 2 , etc. By this 
law of division a 5 -4-a 5 = a°, but a 5 H-a 5 =i because any 
quantity divided by itself is 1 ; therefore, a°= 1, but a is 
any quantity whatever; hence any quantity with exponent 
o equals 1. 

To Multiply a Polynomial by a Monomial. 

Art. 21. To multiply a polynomial by a monomial, mul- 
tiply each term oj the polynomial by the monomial and take 
the algebraic sum of the products, thus: 

3 x 2 y 3 - 2xyz + y 2 z 3 - $x 2 y 
2x 3 y 



6x 5 y 4 — 4-x 4 y 2 z + 2x 3 y 3 z 3 — iox 5 y 



~5*,2 



1 2 Algebra. 

To multiply one polynomial by another, multiply each 
term of one polynomial by each term of the other, and take 
the algebraic sum oj the products, observing the laws of signs 
and of exponents, thus : 

2a 3 xy + aby — $x 2 y — 4xyz 

2 ax — x + $x 2 yz 

4a*x 2 y + 2a 2 bxy — 6ax 3 y — 8 ax 2 yz — 2a 3 x 2 y — abxy+ $x 3 y 
+ \x 2 yz + 6a 3 x 3/ fz + $abx 2 y 2 z — <)x A y 2 z — i2x 3 y*z 2 . 

Again, a 2 - ab + b 2 

a + b 
a 3 -a 2 b+ ab 2 
+ a 2 b-ab 2 + b 3 



a 3 + b 3 

These rules are plain, since a polynomial is the alge- 
braic sum of a number of terms. 

Division of a Polynomial by a Monomial. 

Art. 22. To divide a polynomial by a monomial, divide 
each term of the polynomial by the monomial, dividing the 
numerical factor in each term of the dividend by the numeri- 
cal factor in the monomial divisor, subtract exponents of lit- 
eral factors, and take the algebraic sum of the quotients. 
Thus, 4x 2 yz — 6x 3 y — Sx 2 y 2 + ioxy 3 + 2xy = 2xz — ^x 2 — 
4xy+5y 2 . 

Division of one Polynomial by Another. 

Art. 23. Since the quotient in division is always the 
quantity by which the divisor must be multiplied to equal 
the dividend, it follows that if the divisor and dividend 
are both arranged with the terms having the highest 
power of the same letter standing first, then the first term 
of the quotient will be that quantity by which the first 



Algebra. 1 3 

term of the divisor must be multiplied to equal the first 
term of the dividend, and hence it must be one term of 
the quotient, because this highest degree term of the 
dividend could only be gotten by multiplying the highest 
degree term of the divisor by this term of the quotient. 
Now if every term of the divisor be multiplied by this 
first term of the quotient, the product will be at least a 
part of the dividend. The remainder, obtained by sub- 
tracting this product from the whole dividend, will repre- 
sent what is left undivided of the dividend. If this undi- 
vided part be treated in the same way, the next remainder 
(if there be one) will be another and smaller undivided 
part of the quotient. Plainly each step reduces the re- 
mainder to a smaller and simpler expression, and eventu- 
ally, if the division can be exactly performed, there will 
be no remainder. The following illustration, although 
not exactly analogous, may help to throw light on the 
reason why division is performed as above. Suppose a 
barrel of apples of unequal size is to be divided into three 
parts as equally as possible. The natural procedure 
would be to divide the largest apples into three parts, 
then the next size into three parts, and so on until the 
contents of the barrel is exhausted, that is, there is no 
remainder. 

Example. Divide 3 n A - 25 n 2 — 13 n — n» 3 - 2 by 
1 + 4 n + 3 n 2 . 

Rearranging accord- 3 « 4 ~ I I tt 3 - 25 W 2 - 13 n - 2 I 3 tt 2 + 4 W -f I 



ing to powers: ~ 


ri 


1 + 4 n 3 + n 2 n 2 — 5 n - 2 


Dividing first term of 
dividend by first term 




- 15 n 3 — 26 n 2 - 13 n 

— 15 n 3 — 20 ?r - 5W 


of divisor, multiplying 
and subtracting : 


— 6 n 2 — 8 n — 2 

— 6 n 2 — 8 n — 2 


Again, divide ■ 




25 x 3 y 2 +12 x 4 y +12 x 2 y 3 by 3 x 2 y 


- 4 xy 2 . 







14 Algebra. 

Rearranging according to descending powers of x, 
[according to powers of y would serve as well] and 
dividing : 

12 x 4 y — 25 x 3 y 2 +12 x 2 y 3 k x 2 y — 4-xy 2 
12 x 4 y — 16 x 3 y 2 4 x 2 — 3 xy 

— 9 x 3 y 2 +12 x 2 y 3 

— 9 x 3 y 2 +12 x 2 y 3 



EXERCISE III. 

Multiplication. 
Multiply : 

1. a - b + 2 by 3 a + b. 

2. a 2 - ab + b 2 by a + b. 

3. x 2 — xy + y 2 by x 2 + xy + y 2 

4. 3 ww — 2 ra 2 + w 2 — 1 by m 3 — n 3 . 

5. I a 2 6 - J a + f a& 2 by % a - J 6. 

6. (x + a) (x + &) by (x + c). 

7. x 2 — 2 x + 1 by x - 1. 

8. a 5 - 5 a 3 - 4 a + 8 by a 3 + 2 a - 3. 

9. — 2 a 3 w + m 2 — 3 w 3 + 5 by 3 — 2 a. 

10. 3 x 4 - 4 x 2 y 2 - 2 xy 3 + 5 y 4 by 2 x 2 - 37 + 3 y 2 . 

11. x ?n + 2 x m y M + y n by x a + y 2 . 

12. x 71 - 1 + x M ~ 2 + x n ~ 3 + x n ~ 4 by x - 1. 

13. y* - 2f + 5 xy - x § by x* + 2 x^y* - ?*• 

14. x% - #*y* + y* by x* + y*» 

15. a - aV + 6 by a + aW + b. 

16. 3 (w + ri) h - 2 (a + £)* by (m + w)* + (a + &)*• 

17. a~i + 3 a~i c~^ — 2 c~ x by a* — c~ . 

18. 15 x 4m - 19 x^y™ - 30 x 2m y 2n + 42 x"y w 4- 75 y 4 " 

by 2 x w + y n . 



Algebra. « 1$ 

19. Simplify (x 4 + 3 x 3 - 2 x + 5) (x 3 - 2 x 2 4- 7 x 

-3). 

20. Simplify [(a + b) + 2 c] [(a + b) - 2 c\. 

EXERCISE IV. 

Division. 
Divide : 

1. x* - x 2 y 2 + y 4 by x 2 + xy + y 2 . 

2. x 3 - y 3 by x - y. 

3. w 5 + n 5 by w + n. 

4. 6 (a - 6) n - 2 - 9 (a - ^) n ~ 1 + 12 (a - b) n by 3 
(a - b) n ~ 3 . 

6. a 2 x 4 + (2 ac - b 2 ) x 2 y 2 + c 2 y 4 by ax 2 + bxy + cy 2 . 

6. 8w 4 - 22 w 3 » + 43 w 2 » 2 - 38 w» 3 + 24 « 4 by 2 w 2 
- 3 ww + 4 w 2 . 

7. a 5m + b 3 " 1 by a m + b m . 

8. 9 x n_4 + 19 x"" 1 + 5 x n ~ 2 - 30 x n + 4x n + 1 by 2 

9. 8 a 3 6 3 - 64 x 6 / by 2 afi - 4 x 2 y 2 . 

10. x 2 + y 2 + z 2 + 2 xy + 2 xz + 2 yz by x + y + z. 

11. 1 - a 6 by i + 2 a + 2 a 2 + a 3 . 

12. x - y by x* - y • 

!3. x .+ 3 x* y* - 2 4 x* y* — 3 x* y* - y by x* + 3 x* y* - y*. 

14. What number must be multiplied by x + 2 y 4- 3 z 
to give 9 x 2 + 24 xy + 1 2 y 2 + 30 xz + 24 yz + 9 z 2 ? 

16. Divide 5 rV - 26 rV +2 r 7 j - 5 rV — 11 r 5 s 3 
4- 7 r 2 5 6 - 12 rs 7 by r 4 - 4 r 3 5 + rV - 3 rs 3 . 



CHAPTER II. 
FACTORING. 

Article 24. A factor of a quantity is a divisor of that 
quantity. 

A quantity has as many factors as it has distinct 
divisors. # 

Art. 25. A factor of a factor of a quantity is a factor 
of the quantity itself. 

Illustration : a 4 - b 4 has the factor a 2 - b 2 , and 
a 2 - b 2 has the factor a — b ; hence, a — b is factor of 
a 4 — b 4 , hence : 

Rule — Find the most evident 'factors of a quantity and 
examine these factors for factors. Every step simplifies the 
process. 

Art. 26. There are several general types of quantities 
with respect to factoring. First : The difference of the 
squares of two quantities, which is always factorable into 
the sum and the difference of the two quantities, thus : 

a 4 - b 4 = (a 2 ) 2 - (b 2 ) 2 = {a 2 - b 2 ) {a 2 + b 2 ) ; 

x y _ a 2 * = ( x *y)2 _ (^2)2 = (^ _ ab 2) ( x 3 y + ab 2) . 

(m 2 +2«) 2 - (rs - c) 2 = (m 2 + 2 n + rs - c) (m 2 + 2 n 
- rs + c), etc. 

Second : Trinomials of the type x 2 + ax + b when b is 
the product of two factors whose algebraic sum is a, (the 

* Factors (or quantities in general) are called prime when they have 
no factors except themselves and unity. 



16 



Factoring. 1 7 

signs of the factors being taken into consideration), thus : 
x 2 - x - 6 = (x - 3) (x+2) for b = - 6 = ( - 3) x ( + 2) and 
a = — 1 = - 3 + 2, hence : 

Rule for factoring trinomials of the form x 2 + ax + 6 : 

Separate the last term into two factors whose 
algebraic sum will be equal to the coefficient of 
the middle term with proper sign; each of these 
factors with its proper sign attached to x, will 
form a factor of the trinomial. 

Art. 27. If the trinomial is of the form mx 2 + nx + p, 
it may be factored if m and p are each divisible into two 
factors, such that the sum of the products of these factors 
multiplied diagonally equals tn, thus : 

6= <Bx? 
6 x 2 - 5 x - 4 = (3 x - 4) (2x + 1) for - 4 = - ^ x JL and 
(3Xi)+ (- 4 X 2) = - 5. 

Article 27 may be reduced to the form of Article 26 by 
a simple transformation, thus : 

Multiply 6 ^; 2 - 5 x - 4by6 = 36x 2 ~3ox- 24 = 
(6 x) 2 — 5 (6 x) — 24. Let 6 x = v, then 36 x 2 — 30 x 
- 24 = y 2 - 5 y - 24 = (y - 8) (y + 3) [by Article 26] 
= \6 x - 8) (6x + 3). 

., 6x2 _ 5 x _ 4 _ ( 6 *- 8 H 6 * + 3) = (6 * ~ 8) 

6 2 

X < 6 « + 3) ■ (3 ,. 4) (, x+l) . 



Art. 28. The trinomial that is a perfect square is 
evidently a special case of Article 26. 

Art. 29. It is often possible to factor an expression by 
grouping the terms and removing a factor from similar 



1 8 Algebra. 

groups, thus revealing a common factor in each group ; 
thus : 

x 3 -y 3 -x 2 + y 2 -x 2 + 2 xy - y 2 = x 3 - y 3 - (x 2 - y 2 ) - (x 
- y) 2 = (x-y) (x 2 + xy + y 2 - x - y - x + y) = (x - y) [x 2 
+ xy + y 2 - 2 x], or^V - 8 y 3 z 2 - 4x 3 ?y+ 32 y 3 n 2 = z 2 (x 3 

- 8 y 3 ) - 4 n 2 (x 3 -Sy B ) = z 2 (x 3 - (2 y) 3 ) - (2 n) 2 (x 3 

— ( 2 y) 3 ) =[Z 2 ~ (2 W) 2 ] [^ 3 - (2 )>) 3 ]=(3-2W) (z + 2 W) 
(# - 2 ^) (x 2 + 2 XV + 4 ^ 2 ) . 

A little ingenuity in arrangement and grouping often 
reveals concealed factors, thus : 

x 4 +2 abx 2 -a*+ 3 a 2 P- b* may be written (x*+ 2 abx 2 
+ a 2 b 2 ) - (a 4 - 2 a 2 b 2 + b 4 ) = (x 2 + ab) 2 - (a 2 - b 2 ) 2 - (x 2 + aft 
+ a 2 -& 2 ) (x 2 + ^-a 2 +i 2 ). 

Art. 30. The difference of the even powers of two 
quantities is always divisible by the sum or the difference 
of those quantities ; as, 

x 4 — v 4 = (x — y) (x — y) (x 2 + y 2 ) ; or (m 2 + mn) 2 — (x — y) 2 
= [{m 2 + mn) + (x — y)] [(m 2 + mn) — (x — y)]. 

Art. 31. The difference of the odd powers of two 
quantities is always divisible by the difference of the 
quantities ; as, 

x 3 - y 3 = (x - y) (x 2 + xy + y 2 ) or (a + 2) 3 - (b - i) e 
= (a + 2-b+i) [(a + 2) 2 +(a + 2) {b - 1) + (b- i) 2 ], etc. 

Art. 32. The sum of the odd powers of two quantities 
is always divisible by the sum of the quantities ; thus : 

a 5 +b 5 =(a + b) (a 4 -a 3 b + a 2 b 2 -ab 3 +b 4 ) or (2 X -y) 3 
+ (32+ i) 3 = (2 X -y + sz+ 1) [(2x-y) 2 -(2x-y) (32+1) 
+ (32+ i) 2 ]. 

Art. 33. The sum of the even powers of two quantities 
is never divisible by either the sum or difference of the 
quantities. 



Factoring. 1 9 

Such quantities may be sometimes factored by adding 
and subtracting the same quantity ; thus, 

x 4 +4y 4 = x 4 + 4y 2 x 2 + $y 4 - \y 2 x 2 = (x 2 + 2 V 2 ) - (2xy) 2 = 
(x 2 + 2y 2 + 2xy) (x 2 + 2y 2 - 2xy) etc. 

EXERCISE V. 

Factor the following : 



1. 


x 8 - x 4 y 4 . 


15. 


25 40 + 16 

m 2 wx 3 # 4 


2. 


tf _ b 3 _ IIQ 


16. 


x- 6 - y~\ 


3. 


6u 2 — 2311 + 20. 


17. 


15/ + 5/5-5-3. 


4. 


(a + b) 3 + (a - b)\ 


18. 


a! — wi 


5. 


x 6 + y\ 


19. 


6jc 3 — 7 ax 2 — 20a 2 x. 


6. 


x 8 + y 8 + x 4 y 4 . 


20. 


x 2 + ax + x + a. 


7. 


mb + z 2 — mz — bz. 


21. 


(x + i) 2 - sx - 29. 


8. 


m 2 - 14m— 176. 


22. 


(x 2 + /-z 2 ) 2 -4^y. 


9. 


y 2 - z 2 + 2z - 1. 


23. 


z 5 +7Z 3 - 5Z 2 - 35. 


10. 


f+ y-72. 


24. 


m 3 + m 2 — *jm~ 3 . 


11. 


{2x - sy) 2 - (x— 2y) 2 . 


25. 


4 (z*-^) 2 -(a + 6) ? . 


12. 


n 2 — 2mn + m 2 — x 2 . 


26. 


y 5 + zf - f - 3y- 


13. 


a 2 + b 2 - c 2 

1 + 1 
2 a£ 


27. 


yip + 41/ -45- 


14. 


x 3 - y 3 - (x 2 - f) 
~(x- y) 2 . 


28. 


a 2 - (b - c) 2 . 



GREATEST COMMON DIVISOR. 

Art. 34. Definition : The greatest common divisor or 
greatest common factor (abbreviated G.C.D. or G.C.F.) 
of two or more quantities is the greatest quantity that will 
divide them all. 

Hence, to find the G.C.D. , separate the quantities into 
their prime factors (what are prime factors ?), select the 
factors that are common to all, repeating each factor 
the least number of times it is contained in any one of the 



20 Algebra, 

quantities. The product of these common factors thus 
repeated is the G.C.D. 

For example, find the G.C.D. of 

16 x 2 / z 3 m 3 , 169 y 4 z 6 m, and 39 x 7 y 8 m*. 
16 x 2 yh 3 m 3 = 2.2.2.2. x.x. y.y.y. z.z.z. m.m.m. 
169 y* z Q m = 13.13. y.y.y.y. z.z.z.z.z.z. m. 
39 x 7 y 8 m 4 = 3.i3. 

x.x.x.x.x.x.x. y.y.y.y.y.y.y-y. m. m.m.m. 

The common factors are y and m, y being repeated 
three times as the least number, and m occurring but once, 
hence, G.C.D. = y 3 m. 

Again, find the G.C.D. of 

x 2 + 5X + 6, x 3 + yx + 10, and x 2 + \2x +20. 

x 2 + $x + 6 = (# + 2) (# + 3) 

x 2 + jx -f 10= (x + 5) (# + 2) 

X 2 + I2X + 20 = (x + 2) (x + 10) 

x + 2 is the only common factor, therefore, x + 2 = G.C.D 

G.C.D. Without Factoring. 

Art. 35. Let A and B be any two quantities of which, 
say, A is the greater, and let B be contained in A , Q times 
with a remainder 7?, then, 

^ = QB + R or R = ^ - ()£. 

Since the sum of a fraction and a whole number can 
never equal a fraction, any factor common to A and B 
must be contained in R ; otherwise, if we were to divid? 
the above equation by such a factor, the quotient on the 
right of the equality sign would be a whole number (since 
the factor exactly divides A and B), and on the left it 
would be a fraction, which is impossible ; hence, R con- 
tains all the factors common to A and B. But it may 



Factoring. 2 1 

also contain other factors. If we divide B by R and say 
R is contained in B, M times, with a remainder R f , then 
B = MR + R f . By the same reasoning R f contains all 
the factors common to R and B, and hence all factors 
common to A and B. 

Suppose finally that a remainder obtained by such 
successive division is contained exactly in the previous 
remainder, then it represents all the factors common 
to A and B, and no others ; hence, it is the G.C.D. of A 
and£. 

Hence the rule : Divide the greater quantity by the 

LESS AND IF THERE IS A REMAINDER DIVIDE THE FIRST 
DIVISOR BY THIS REMAINDER, AND THEN THIS REMAINDER 
BY THE REMAINDER RESULTING FROM THIS LAST DIVISION, 
AND SO ON UNTIL A REMAINDER IS EXACTLY CONTAINED IN 
THE PREVIOUS REMAINDER. THIS LAST REMAINDER IS THE 

G.C.D. 

Example. Find the G.C.D. of 



v 3 — 3X 2 + $x — i 
x + I 



x 4 - 


2x 3 + 2x 2 —2x+ 1 and 


x 3 — 3.V 2 + $x - 1 


X 4 — 2X 3 + 2X 2 — 2X + I i 

x A — 3.V 3 + 3V 2 — X I 


x 3 - X 2 — X +1 


x 3 - $X 2 + $X — I 


2 


2X 2 — 4X +2 



X" — 2X +1 

Since the G.C.D. contains only common factors, the re- 
moval from or introduction of any factor into either one, 
if it is not also a factor of the other, does not affect 
the common factors, and hence does not affect the G.C.D. 
Hence, we may take out the factor 2 from above remainder, 
without affecting the result. 



22 Algebra. 



To continue : 






x 3 - 2>x 2 + 3x - i 


|* 2 - 


■25(7+1 


x 3 — 2X 2 + X 


# - 


I 


— X 2 + 2X — 1 






— X 2 + 2X — I 






Hence, x 2 — 2 x + 1 = G.C.D. 





Again, find the G.C.D. of 

# 5 — $x 4 - 3# 3 — 15 — 195; and 

Sx 4 — $x 3 + x 5 — 15 + q# 2 — #. 

To facilitate the process, both quantities should first 
be arranged according to the ascending or descending 
powers of one of the quantities involved. 

Rearranging the above and dividing: 

x 5 -$x 4 -3x 3 - igx - 15 jv 5 + 3.V 4 - 3X 3 + gx 2 - x - 15 
x 5 + 35c 4 - 3X 3 + gx 2 - x - 1 5 1 
— $x \—6x 4 +9x 2 — i8jc (Remove factor — 357 which 
2x 3 — 2>oc + 6 does not occur in the divisor.) 

Multiply first divisor by 2, since 2 is not a factor of 
2a7 3 -3:x7+6. 

2# 5 + 6x 4 — 6x? + i8# 2 - 2x - 30 \2x 3 — 35; + 6 

5c 2 + 35; — 3 



make divisible.) 
(Divide by 7.) 



2X 5 


— 3.x 3 + 6x 2 




6x 4 — $x 3 + 1237 2 - 2x — 30 
6x 4 — 9^; 2 + i8x 




— 3X 3 + 2I# 2 — 20JC7— 30 
2 




— 6x 3 + 42X 2 — \ox— 60 

— 6x 3 + gx— 18 




7 42 x 1 — \gx- 42 



6x 2 — jx — 6 



Factoring. 23 

Multiply {2.x 3 - $x + 6) by 3. 

6x* - gx + 18 fix 2 - 7s - 6 

x + 7 



6X 3 - 


yx 2 - 


6* 


7# 2 - 


3X + 


18 
6 


42X 2 — 
42JC 2 - 


i8x + 
49* - 


108 
42 



3IX + 150 

which is plainly not contained in by? — jx — 6 and 
hence the original numbers had no G.C.D. 

It is unnecessary to carry the process further, as the 
next division would leave a remainder containing only one 
known term, which could not be the G.C.D. of quantities 
containing unknown terms (as x). 

To Find the G.C.D. of More than Two Quantities. 

Art. 36. Rule. Find G.C.D. of any two of the quanti- 
ties and then the G.C.D. of this G.C.D. and a third quantity, 
then the G.C.D. of this last G.C.D. and a fourth and 
so on until all the quantities have been used. The last 
G.C.D. will be the G.C.D. of all the original quantities. 

EXERCISE VI. 
Find the G.C.D. 

1. 8 - a 3 ; a 2 - 4. 

2. x* - v 6 ; x 4 + xy 3 ; x* + 2x*y* + v 6 . 

3. f - 3y + 2 ; v 4 - 6f + 8v - 3. 

4. u 3 — $u 2 + 4; 3U 3 - iSu 2 + 3611 — 24. 

6. (a + b) 2 ~ (c + d) 2 ; ax + bx + ex + dx. 

6. x 2 + 5x + 6 ; x 2 + jx + 10 ; x 2 — x — 6. 

7. a 3 - b\ (a - b) 3 ; a 2 - 2ab + b 2 . 

8. z 3 + 2z 2 + 2z + 1 ; z 3 - 4Z 2 -42 — 5. 

9. m 3 - 19W - 30; m 3 + 10m 2 + 31W + 30. 



24 Algebra. 

10. x 4 — 4.T 3 — i6x 2 + jx + 24 ; 2x 3 — 15X 2 H- 9^+ 40. 

11. 6x 2 + x - 2 ; gx 3 + ^&x 2 + 52^ + 16. 

12. 2 / 4 - 3/ 3 - 9* 2 + 9/ - 2 ; 3/ 5 - 4/ 4 - 23^ + 4^ 2 

- 20/ + 3. 

13. x 4 — 9X 2 — 30X — 25 ; x 5 + x 4 — *jx 2 + $x. 

14. 2x 2 — 7x + 3 ; 3.T 2 — Jx — 6. 

15. y *- 2 f- 13/ + 383; -24; /- 4/ -7/ + 34^- 24. 

16. $x 3 + 9x 2 y—6xy 2 — 6y 3 ; 2^x 3 + 6x 2 y— 12 xy 2 — iSy 3 . 

17. iow 3 + n 2 - gn + 24; 2on 4 - ijn 2 + 48^ — 3. 

18. 12 (a 4 - b 4 ) ; 10 (a 6 - b 6 ) ; 8 (a 4 6 + a& 4 ). 

19. x 3 - 3X 2 - 4x + 12 ; x 3 - 7^ 2 + i6x - 12 ; 2x 3 

- 9X 2 +7^ + 6. 

20. x 2 + 1 ix + 30 ; 2x 2 + 21.T+ 54 ; 9# 3 + 53x 2 -gx- 18. 

21. z 2 + us + 30 ; z 3 - i2z 2 + 412 - 30 ; z 4 - i2z 3 
+ 472 2 -72z + 36. 

22. 6X 2 + x - 2 ; gx 3 + 48X 2 + 52X + 16. 

23. 2w 4 - 5 n 3 - 3 w 2 ^ 2 + 7 w/> 3 + 3 p 4 ; 8w 3 - 4 w 2 /> 

- 8w^ 2 - 6^ 3 . 

24. x 4 + x 2 / + y 4 ; x 8 + .r 4 y 4 + y* ; X 16 + tf 8 / + >' 10 . 

LEAST COMMON MULTIPLE. 

Art. 37. The least common multiple of any number of 
quantities (abbreviated L.C.M.) is the least number that 
contains them all. 

From this definition, the following rule is immediately 
inferred : 

TO FIND THE LEAST COMMON MULTIPLE OF ANY NUMBER 
OF QUANTITIES, SEPARATE THESE QUANTITIES INTO THEIR 
PRIME FACTORS, AND SELECT ALL THE SEPARATE FACTORS 
THAT OCCUR IN ALL THE QUANTITIES, REPEATING EACH ONE 
THE greatest NUMBER OF TIMES IT APPEARS IN ANY ONE OF 
THE QUANTITIES. THE PRODUCT OF THESE FACTORS WILL 
BE THE L.C.M. 



Factoring. 2$ 

Example : find the L.C.M. of 

2jx 7 y, 6x 2 y 4 , 4xy 5 , $x 4 y 4 
2jx 7 y = 3-3-s- x.x.x-x-X'X.X' y- 
6x 2 y 4 =3.2. x.x. y-y-y-y- 
qxy 5 = 2.2. x- y-y-y-y.y. 
$x 4 y 4 = 3. x-x-x'x- y-y.y.y> 

The separate factors are 3, 2, x, and y. 



3 occurs 


3 


times 


in 2jx 7 y. 


2 " 


2 


u 


" 4^ 5 - 


x " 


7 


(( 


" 2 7X 7 ^. 


y " 


5 


(( 


" 4JC^ 5 . 



Hence, L.C.M. is 3 . 3 2. 2 xy = io8^y 



To Find L.C.M. Without Factoring. 

Art. 38. Since the product of two quantities contains 
all the factors of both quantities with the common factors 
repeated as many times as they occur in both together, 
and since the G.C.D. of these two quantities contains 
only the com won factors, if the product of the two quanti- 
ties be divided by their G.C.D. the quotient will contain 
all the factors of both and their common factors once 
only; hence, this quotient is their L.C.M. Put this into 
rule : 

If there are more than two quantities, find the L.C.M. 
of any two of them, then the L.C.M. of this L.C.M. of 
two and a third quantity, then the L.C.M. of this last 
L.C.M. and a fourth, and so on, until the quantities have 
all been used. The last L.C.M. will be the L.C.M. of all 
the quantities. 



26 Algebra. 

EXERCISE VII. 

Find the L.C.M. of 

1. 16 (x 3 - y 3 ) ; 24 (x 4 - y A ) ; 36 (x 3 + y 3 ). 

2. x 2 + jx + 12 ; x 2 + x — 12 ; x 2 + 3X — 4. 

3. (a + 6) 2 -£ 2 ; (a- b) 2 - c 2 ; + c) 2 ; (a- <0 2 -& 2 . 

4. x 2 — $x — 70 ; x 3 — 39X + 70. 

6. 4s 2 - yst + st 2 ; 3s 3 - 4s 2 / + 3s/ 2 - 2* 3 . 

6. a 2 — 3a - 4; a 2 — a— 12 ; a 2 + 5a + 4. 

7. 2^+5-83; + y 2 ; 42y 2 + 30 - 723;. 

8. 3X 4 — x 3 — 2x 2 +2x — 8; 6x 3 +i3x 2 + 3X + 20. 

9. w 5 — 2U 4 + U 2 ] 2U 4 — 411 3 — \u —4. 

10. x 3 — 6x 2 — $x + 1 2 ; x 3 — 5X 2 + 2x + 8; x 3 - 4X 2 + x + 6. 

11. 3Z 2 - 52; 4- 2 ; 4Z 3 - 4Z 2 - z + 1. * 

12. / - 4 P ; f + 2^ + 4% + 8b 3 ; / - 2b f + \b 2 y- Sb 3 

13. a 2 + a — 2 ; a 3 + 2a 2 + 2a + 1. 

14. x 3 — x 2 - gx + 9 ; x 4 — 4X 2 + 12X — 9. 

15. 4Z 3 - 8z 2 + 52 - 3 ; 2z 4 - 3s 3 + 6z 2 -32+2. 

16. x 2 + 5X + 6 ; x 2 + 6x + 8 ; x 2 — 3X — 10. 

17. Find the least quantity, which when divided by y 2 + y 
— 2,y 2 — y—6, and y 2 — \y + 3, leaves in each case the 
remainder, 2V 2 - 3^ + 1. 

18. The G.C.D. of two quantities is x 2 — xy + y 2 , and 
their L.C.M. is x 5 + x 4 y 4- x 3 y 2 + x 2 y 3 + x 2 y 3 + xy 4 + y*. 
One of the quantities is x 3 + y 3 . Find the other one. 

19. Four pendulums beat, respectively, every second, 
every 1 J seconds, every 1+ seconds, every 1 \ seconds. They 
are started together ; after what time will they all beat 
together again, and how many beats will each make in 
the interval ? 



Factoring. 27 

FRACTIONS. 

Art. 39. The rules which apply to operations with 
fractions in arithmetic, apply equally to algebraic fractions, 
if the slight modifications of the four fundamental opera- 
tions, addition, subtraction, multiplication, and division, 
applied to algebraic quantities, are observed. 

It must be remembered that the use of letters in algebra 
often makes it impossible to condense expressions, as in 
the use of numbers exclusively. Hence the operations 
are more frequently indicated than actually performed. 

Although, for example, if really means 1 + f , the 
structure is not so apparent as in the exactly similar alge- 

b 
braic expression, a + — • 

Both are mixed numbers and both are reduced to frac- 
tions in exactly the same way ; viz. 

5 

, b aXc + b ac +b. 

a -\ — = = • 

c c c 

Art. 40. Reduction to lowest terms, reduction to 
common denominator, addition, subtraction, multiplica- 
tion, and division, are accomplished in ways exactly analo- 
gous to the similar arithmetical operations, the algebraic 
concept of sign, and the use of letters, introducing a 
purely superficial difference. 

A general rule may be formulated thus : 

Perform the required operation as indicated in arithmetic, 
observing the laws of algebraic addition, subtraction, multi- 
plication, and division. 

It is to be observed that in reducing an algebraic frac- 
tion to its lowest terms, every term of both numerator and 
denominator must be divisible by the factor removed, as, 



a 2 x + aby 


ax + by a 2 x + aby 


ax + aby 


ac 2 - a 3 d 


c 2 - a 2 d L ac 2 - a 3 d 


c 2 - a 3 d 



28 Al zebra. 



Signs. 

Art. 41. Since changing all the signs of both numer- 
ator and denominator is equivalent to multiplying (or 
dividing) by — 1, the value of the fraction is not changed 
thereby. But if the signs of either numerator or denomi- 
nator (but not of both) be changed, or if the signs of any 
parts (not all) of either or both be changed, the value of 
the fraction is changed. Thus : 

x 2 y — xy 2 xy 2 — x ? y x 2 y — xy 2 

' 1 — i = ~l out —j — ; — 

— abc+ mn a be — mn a be + mn 

x 2 y + xy 2 
is not equal to ~ 7 ' • 

n abc — mn 

Also when a fraction has a minus sign in front of it, the 
sign of each term of the numerator must be changed, 
when the denominator is removed or when the numerator 
is combined with any other expression (as in adding or 
subtracting). The effect being the same as removing a 
parenthesis. Thus : 



ax 



+ b ex + d a 2 x + ab — c 2 x — cd 



c a 



EXERCISE VIII. 


Reduce to lowest terms : 




x 4 + (2b 2 - a 2 ) x 2 

1 

x 4 + 2ax 3 + a 2 x 2 ■ 


+ b 4 
- b 4 


a 2 — a - 12 


4 ^ ~ {C ~ ^ - e 2 

(b + d) 2 C 


2 ° (a + 3 ) 3 


3 y 4 + f - 2 
y 4 + 5 y ' + 6 


6 x 2 + 1 3 x + 6 
* 10 x 2 + 13 *- 3 



Factoring. 29 



Reduce to mixed numbers : 

12 u 2 — 5 w — 6 z 3 - 3 z 2 + 2 z 

6. r-*- — • 8. 

3 u* + 2 z — 1 



ar + x 2 — i m A — n 



7, : 9. 



ar m — n 



Simplify the following : 

2 m — 6 

10. - + 



w 3«+6 ra 2 + 2 m 
2 2 # 



11. 7 TT + 

(a — 1/ # — 1 X 2 + 1 

12 . s 4 7 

# 2 — 2 x — 3 x 2 — 9 x 2 + 4 ^ + 3 

X - 2 (x — 1) [X — 2) s 

ax + I) c 

14. 



X Z + 1 2 X — 3 

16. --*- + -J ** 



x - 1 # — 3 x 2 -4^+3 

4 «4 



\m 2 — n 2 m? - mn) \m — n mn — n 2 ) 



a + b . b + c 

17. ~, r + t 



(b - c) (c - a) {c - a) {a - b) (a - b) (b - c) 

2 x + y y 
18. L ~ 

y 2 x + y 



x _ x + y 
x + y x 

\y 5 x 5 ) "*" \y x) 



20. 1 + — - 



x + i_ 

X 



CHAPTER III. 

EQUATIONS. 

Article 42. An equation is a statement of equality 
between two equal expressions, thus : 

ax 2 + bx + c = d, or $x + 2y + z = 24, etc. 

Art. 43. An equation usually consists of letters and 
figures, although it may contain either exclusively ; as, 

$x + y = 11, or ax + b = c, or 5 + 2 = 7. 

The last is called an arithmetical equation, the others 
algebraic equations. 

Of the quantities involved in an algebraic equation, part 
are usually known, that is, their values are known, and part 
are unknown. It is customary to represent the latter by 
the last letters of the alphabet ; the former by the first 
letters of the alphabet or by figures. 

Thus : x, y, z, etc., are the symbols for unknown value, 
and a, b, c, d, etc., 1, 2, 3, 4, etc., for the known. 

Art. 44. Equations involving only letters are called 
literal equations. Those involving unknowns and figures, 
numerical equations. 

Art. 45. Solving an equation is the finding of the 
value or values of the unknown quantities that make it 
true, or, technically, satisfy it. That is, the finding of the 
values which when substituted for the unknowns to which 
they belong, make the two sides of the equation the same. 
The. equation then becomes an identity. 

Art. 46. An equation is usually satisfied only for a 
30 



Equations. 3 1 

limited number of values of the unknown quantities. Thus : 
$x + 6 = 1 2 is satisfied only when x * ^ 2 , for then, and 
then only, we have 3 (2) + 6 = 6 + 6 = 12, an identity, 

or, ax + b = c is satisfisd only by x = — — , for then 

a 

we have a -^-^ — '- + b = corc-j/+p=c. 
a ' 

x 2 - $x + 6 = o is satisfied only when x = 2 or 3 
for (2) 2 - 5 (2) + 6 = 4 - 10 + 6 = o, or (3)2 - 5 
(3) + 6 = 9 - 15 + 6 = o. 

When an equation is of such a nature that any value 
whatever of the value unknown satisfies it, it is called an 
identity and is written thus : 

ax + b = 2x + 3 

Uses of Literal Equations. 

Art. 47. A literal equation is of special service, be- 
cause the value (or values) of the unknown obtained from 
it is general for all cases of the same kind. For instance, 
it is required to find the mean of any two quantities (that 
is the quantity midway between them). 

Let x = the mean and let a and b represent any two 

quantities ; then a — x = x— b, or 2 x = a — b ; x = . 

2 

Such a result is called a formula, because it is true for all 
quantities, and may be stated as a general rule. For 
instance, in the above case, the result may be expressed 
thus : 

The mean oj any two quantities is half their sum. 

Uses of Letters in General. 

Art. 48. The use of letters enlarges the value of results 
and gives algebra a notable advantage over arithmetic. 
Suppose it is required to prove that half the sum of two 



32 Algebra. 

numbers plus half their difference equals the greater 
numbers. 

If we use a and b as the numbers, a being greater, we 

have 1 = — = a ; which gives a general 

2 2 2 

result, as a and b may stand for any numbers. 

Art. 49. The use of letters is valuable again because 
all the operations performed must be indicated and the 
anatomy of the result is evident, whereas the ability to 
combine simple figures covers up the traces of the opera- 
tions where they alone are employed. Thus, if (x + a) is 
multiplied by (x + b), we get x 2 + {a + b) x + ab, which 
reveals the well-known rule for factoring trinomials. What 
is it? 

While if (x + 2) is multiplied by (x — 5) we get x 2 
— $x — 10 wherein there is no evidence of the relation 
between — 3 and - 10, and the coefficients, 2 and — 5. 



Degree of an Equation. 

Art. 50. The degree of an equation is the largest sum 
of the exponents of the unknown quantities in any one 
term ; for example, x 2 y + 2x + 3 V 2 + xy = o is of the 
third degree, for the first term contains x 2 and y 1 making 
sum of exponents 3, and no other term is higher. But 
a 2 x + aby + ex 2 is of the second degree, for the exponents 
of a, b, and c do not count, as they are regarded as known 
quantities. 

Equations of the first degree like ax + b = c, or $x 
+ Zy = 7» etc -> are called linear equations ; those of the 
second degree like ax 2 + bx + c = o, or $x 2 + 2xy = y 2 , 
are called quadratics ; those of third degree, cubics ; of 
fourth degree, bi-quadratics, etc. 

Art. 51. An equation may plainly contain any num- 
ber of unknown quantities. In order to find the value 
or values of the unknowns, the number of independent 



Equations. 3 3 

equations must equal the number of unknown quantities 
involved, because each unknown requires a separate con- 
dition to distinguish it from the others. 

Art. 52. When two or more equations involving two 
or more unknown quantities, are all satisfied for the same 
values of those unknowns, the equations are called simul- 
taneous equations. 

They are said to be independent when no one of them 
can be derived from any other. They must represent 
entirely distinct conditions. 

Thus : 3X + 2y = 8 

6x + \y = 16 are simultaneous but not inde- 
pendent. 

Which of the following are independent ? 
Are they all simultaneous ? 

1. j 15X + yjy = 92. 2. ( 6y - $x = 18. 

( 5* - 37 = 2 - X i2 * - 9y - °- 

3- j 5* + by - 17. 4. { SP + m = G %- 

1 2/> + 5? =69. 

6. + ** ' 6. A 3 3 



Clearing of Fractions. 

Art. 53. Since the equality of two quantities is not 
affected by multiplying both by the same or equal quantities, 
the fractions may be removed from an equation by multi- 
plying by the L.C.M. of the denominators involved. 
Thus : 

%x 2 +%x+% = % may be cleared of fractions by mul- 
tiplying both sides by 70, the L.C.M. of the denominators, 
and the result, 105 x 2 + 20 x + 42 = 35, is free of 
fractions. 



i 


5* 


+ 


6y = 


17. 


( IOX 


+ 


I2V = 


34- 


'X 


+ 


Z 


= 12. 




4 




2 






X 

,2 


+ 


y 


= 27. 





34 Algebra. 

The L.C.M. of all the denominators is called the Least 
Common Denominator, abbreviated to L.C.D. 

Note. Of course any multiple of the denominators will 
remove fractions, but the L.C.D. gives simplest result. 

Transposition. 
Art. 54. Rule. Any term may be transferred from 

ONE TERM OF AN EQUATION TO THE OTHER IF ITS SIGN BE 
CHANGED. 

Let x + b = c be any equation ; if — b be added to both 
terms the equality is not affected, then x + X — fl = c —b 
or x = c — b, which justifies the rule. Again, let the 
equation be x — b = c, and add + b to both sides, x — ty 
+ h = c + b ox x = c + b, which again verifies the rule. 

To Solve a Linear Equation of One Unknown. 

Art. 55. Since to solve an equation is to find the value 
of the unknown that satisfies it, the requisite steps are to 
the end of simplifying the equation to the utmost. Hence : 

Clear the equation of fractions, if necessary; 
transpose all the terms containing the unknown to 
one side of the equation (preferably to the left 
side), and all other terms to the other side, col- 
lecting them; combine the terms containing the un- 
known by collecting where possible and by express- 
ing the result as the product of two factors, one of 
which is the unknown itself; divide through by the 
other factor, and the result is the value of the 
unknown, or the solution of the equation. 

EXAMPLE. 5X ~ 6 - l£ = ^-^2 (L.C.D. = 20). 
5 4 10 

Clearing of fractions : 20 x — 24— 15 x= 2 x— 18. 

Transposing: 20 x — 15 x— 2 x = 24— 18. 

Collecting : 3 x = 6 or 3 (x) = 6. 

Dividing by 3 : x = 2. 



Equations. 3 5 

Again i?^-^ - I3 f~ 2 f =3 (L.C.D.=^- 9 a 2 ). 
x + 3 a x z — 9 or 

Clearing : x 2 - 5 a + 6a 2 - (13 a 2 - 2 x 2 ) = 3 x 2 - 27 a 2 . 

Removing parentheses : * 

# 2 - 5 ax + 6 a 2 - 13 a 2 + 2 x 2 = 3 x 2 — 27 a 2 . 

Transposing : 

x 2 +2x 2 -2>x 2 -$ax=-6a? + 13 a 2 — 2 7 a 2 . 
Collecting: 

- 5 ax= — 20 a 2 , or (- 5 a)x = — 2oa 2 \x = 4 a. 

Hence the rule : When a parenthesis, before which is a 
minus sign, is removed, or the denominator of a fraction, 
before which is the minus sign, is removed, the signs of 
all terms included in the* parenthesis or in the numerator 
must change. 

Simultaneous Linear Equations. 

Art. 56. Such equations must be solved by finding a 
single equation, involving but one unknown quantity, which 
includes all the conditions. Just as combining any two 
expressions (by multiplication, addition, or otherwise) in- 
troduces into the result all the factors and other charac- 
teristics contained by both, so the combination of two or 
more equations gives an equation which contains the 
qualities and conditions of all. Upon this fact is based 
the rule for solving simultaneous equations. 

* Since a minus sign always means a subtraction, actual or indi- 
cated, the minus sign before the parenthesis in the above example 
indicates that the quantities within are to be subtracted, and since 
removing the parenthesis partially performs the operation, according 
to the law of negative numbers, the signs of the quantities inclosed 
must all be changed. The same thing applies to the numerators of 
fractions, when the denominators are removed by clearing. 



36 Algebra. 

Example. 4 #+9 3/ =79 (Are these independent?) (1) 

7^-17^ = 40 (2) 

Multiply (1) by 7 and (2) by 4; then (1) by 17 and (2) 
by 9. 

28./+ 63 y = 553 68 x + 153^= 1343 

2fx — 68 y = 160 63 x — 1, 33 y = 360 
Subtract, J 3 i y = 393 Add 131 # = 1703 

Divide by 131, y = 3 x = 13 

The equation 131 y = 393, while it contains only the 
unknown y, by the above principle contains all the con- 
ditions of both original equations ; hence, the value 3 of y, 
found from it, will agree with the value of x (13), similarly 
found, and will satisfy with it both equations. 

Verification : 

4 (13) + 9 (3) = 79 or 5 2 + 2 7 = 79> that is 79 = 79- 
7 03) ~ I 7 (3) = 4o or 91 - 51 = 40, that is 40 = 40. 

Art. 57. The process of combining simultaneous 
equations, so as to reduce the number of unknown quanti- 
ties in the result, is called elimination. 

There are several devices employed to effect elimination, 
one of which is illustrated in the example solved above. 
This method may be thus described : 

Multiply each equation if necessary by a factor 
(preferably the least one), that will make the co- 
efficients OF ONE of the unknowns the same in both. 
If the terms containing these unknowns have the 
same sign, subtract one equation from the other; if 
they have opposite signs, add. 

If there are more than two equations, say three, combine 
the equations in pairs (say 1st and 2d, and 2d and 3d), 
thus acquiring two simultaneous equations containing only 
two unknowns. The combination of these two will elimi- 



Equations. 37 

nate one of the two remaining unknowns, leaving but one. 
Then divide by the coefficient of this remaining unknown, 
and the result will be its value for the set of simultaneous 
equations. 
Example. 

2 x + 3 y = l2 ~ \* or 2 x + iy + i* = i2 (*) 

± y+ i z= 8 + ixor-ix+±y + iz=S (2) 
ix + Jz = 10 \x + Jz = io (3) 



ix+iy+ T \z= 6 

Subtract J of (2) from ^ of (1) — T * g ac + jy + ^z 



Multiply (4) by 12 add (3) to (4) 



/ 



11 r 1 r, — 

36 X ""56 Z_ 


¥ 


\) v-*-**- 


40 


%x+§z= 


10 



(4) 



V* =50 
x =12 

Substituting x = 1 2 in (3) 6 + ^2=10, 2=12 

Substituting jc= 12, z= 12 in (1) 6 + Jv+2 = i2,v=i2 



Elimination by Substitution. 
Art. 58. 

Example. A = — ^— . . 

z* 2-7; 



CO 



1 2 

« - 1 t; + 2 
From (1) 6-31/ = u. 

_6 — u 
3 

Substituting value of 

6 — u • / x 1 * 

V - -m (2) = -i- + 3 .... (3 

3 w — I 6 — M 



38 Algebra. 

Simplifying (3) — — = ~ — or — — = 

u— 1 6 — U + 9 u—i 15 — w 

3 

Clearing of fractions, 

15 — u = 6 u — 6, 
whence —7 w= — 21, w = 3. 

From (3) if si « 3, v = — — ^ = 1. 

3 

Rule. Solve one of the equations for one of the 

UNKNOWNS IN TERMS OF THE OTHER AND KNOWN TERMS ; 
SUBSTITUTE THIS VALUE FOR THE SAME UNKNOWN IN THE 
OTHER EQUATION; CLEAR OF FRACTIONS (iF NECESSARY) J 
COLLECT AND DIVIDE THROUGH BY THE COEFFICIENT OF THE 
UNKNOWN INVOLVED. THE RESULT IS A ROOT OF THE 
ORIGINAL EQUATIONS. THIS ROOT SUBSTITUTED IN EITHER 
ORIGINAL EQUATION GIVES THE VALUE OF THE OTHER 
UNKNOWN. 

Elimination by Indeterminate Coefficients. 
Art. 59. 
Example: $ x + 5? = 1 9 (0 

5X - 4 y = 7 ->"(*) 

Since either equation may be multiplied by any factor 
whatever, we will multiply (2) by, say, m, and add. 

3x+5y=ig (1) 

$tnx — \my = ym . . . . . . (2) 

$x + $mx + Sy — 4-my = 19 + 7m 

or x (3 + 5m) + v(5-4w)=i9 + 7m . . (3) 

Equation (3) plainly contains both (1) and (2) and the 
conditions that belong to both, hence, whatever is true of 
(3) is true of both (1) and (2). 

Since m may have any value, let it be — f, so that 



Equations. 39 

the coefficient of x shall be zero, which is plainly equiv- 
alent to eliminating x from (i) and (2), as previously 
explained. Then : 



or 



Substituting 




m = 


3 

5 


in (3) 


x (3 - 3) + 


y(s + 


f)- 

^ y = 

y = 


19- 

74 

5 

2. 


21 

T 


Again, let 




M = 


| in 


(3), then: 


*( 3 + f/ 


) + y(.s 


-5)- 

37* = 

4 

# = 


19 

in 
4 
! 3- 


fJ 35 

4 


Rule. Multiply 


EITHER 


EQUATION 


BY ANY LITERAL 



factor; add (or subtract) this result from the other 
equation, and in the resulting compound equation 
give m successive values that will cause the coeffi- 
cients of the unknowns to successively vanish j the 
value of the remaining unknown will be the value 
that satisfies the simultaneous equations. 

More than Two Unknown Quantities. 
Art. 60. 

(jx + sy- 2s = 16 . . . . . (1) 

Example. -j 2x + 57 4- 32 = 39 (2) 

U#- ? + 5z = 3 1 (3) 

Multiply 2 by (a) and 3 by (6) and add to (1) 

Jx + 2>y ~ 2Z = J 6 
2a# + 5a)/ + 3a:*: = 39a 
56. v - by+ 562 = 316 

(7 + 2a + 56) * + (3 + sa - b)y 4- (3a 4- 56 - 2)2 = 16 
+ 390 + 316 (4; 



40 Algebra. 

We must evidently determine the values of a and b that 
will simultaneously reduce two of the terms of (4) to zero, 
say the x and y terms. That means that we must solve 
the two equations. 

7 + 2a + 5& = o for a and b . . . (5) 
and 3 + $a - b = o ....... (6) 

Transpose and multiply (6) by 5 



Add 



Whence substituting 

22 • , v 44 , , 

a= -in (5), 7- ^i+ 56 =0. 

27 27 

Whence 

27 27 

Substituting this value of a and b in (4) 

27 27/ 27 27/ 

858 899 _ 
27 27 

_L_5_ z = _ _3£5 5 z = <. etc 



2a + 


5* = 


-7 


25a- 


5^ = 


-i5- 


27a 


= 


— 22 


a 


= 


22 
27 



/_£i__I45 _ 2 \ 2=l6 
V 9 2 7 / 



27 27 

Art. 61. State the rule for elimination of any number 
of unknown quantities. 



Equations. 4 1 

General Rule. Combine the equations two and two, 

ELIMINATING ONE OF THE UNKNOWN QUANTITIES; THEN 
COMBINE THE RESULTING EQUATIONS TWO AND TWO, ELIMI- 
NATING ANOTHER UNKNOWN, AND CONTINUE UNTIL THERE 
REMAINS BUT ONE EQUATION CONTAINING ONE UNKNOWN, 
THE VALUE OF THIS LAST UNKNOWN WILL BE ONE OF THE 
VALUES SOUGHT. SUBSTITUTION WILL GIVE THE OTHERS. 

EXERCISE IX. 

1. (y+4) (y-2) = (y + 3) (3y+4)-(2y+i) (y-6). 

2. (x + 2) 2 + $x (x - 2) 2 + 5 (16 - x). 



3. 



x — 2 _ x — 5 

2X— $ 2X — 2 



A 2 S IO 

4. 2 = o. 

2-2 32-4 



5. 



x — 3 12 — 2jc _ 3-v — 27 
x 2 - 9 x l — 36 x 2 — 8 1 



6. =x. 

X 2 + 2&X + b' 



7. (> + ^Z_) _<>__»*_)..,. 

;y — a y — a 

8. 7z-9 = 92 + 4 _ 6z - 5 ^ 
2-3 3 6 2 4 

9. 2X — * - = — + I. 

I I 2 

10. aft* + ^ = ac+ ™. 

bx b 

11. 6 ^ + 7 + 7y - 13 = 2 X±_±. 

9 6 ^ + 3 3 



42 



Algebra. 



12. 



13, 



2X + I 2X - I 



9^ + 17 



2^—l6 2X + 12 X 2 — 2X — 48 

•3 -7 .2-3; 



14. 1 + 



.OI — .02X .OI + .02X .OI (4.X 2 — i) 

3 
-3 =10. 



1 + 



1 — x 



15. 



16, 



17o 



18. 



19. 



20. 



3j(57jz_3i = 1. 

2 (Ay + 3) 5 " 

x -3 + 2 ^ - 1 = j 
2X + 1 4X — 3 



x ~ 3 + * = 2Q _ Jg ~ 19 



2—2 Z + 2 2T — 4 

ax — b bx+ c 



abc. 



21. f 



c a 

6y + 7 _ 3 = 

3 y + 2 

fx + I _ I 



2^+J. 



22. 

23. 

24. 



i + 



f +x f + x 
2 _ 13 5? 



8 4j + 8 



X — 7 _ 2X — 15 _ I 

x + 7 2x - 6 2 (x + 7) 

(3a — #) (a — 6) + 2ax = 46 (a + x). 



Equations. 



43 



EXERCISE X. 

Equations containing more than one unknown, 
1 (Sx = 2y + 78 



+ u = 15 
-f v = 6. 





<3 y = 


= x + 104. 


3 y 

4+ z = 4 




3. 


3 


y-8- 5 






11 — V 

< 5 



4. 



2 ^ — rv 2^—59 

_ = 20 _j_ 

23 - * 2 

73-3 y . 



y - 



3° - 



x y x 

x y y 

- + ~ = 1 + • 
Vb a c 

5 / + 4 * = 49t 
2 / + 7 5 = 63. 

7 + .x 2 .v 



2 x — y 



2 y + x 



+ 14 = i« 



+ 16 



19. 



9. J 



5 4 

5^-7 . 4* - 3 



+ 



3 x + 16 v = 5 

- 5 .v 28 y = 19. 

= 3 y - 5 
18-5 ». 



10. 



4? + 



3 y = 43- 



y z 

This means — 4- — 

2 3 



->•] 



11. 



U V 

9 IO 

•u V 



12. 



3 y y 



12 .v - 3 v y - 2 



44 



Algebra. 



15. 



16. 



18. 



20. 



x — y 



— _ 7 

4 



13. 1 1 — x — y 

■2 x + 3 y = -- i 



* - 3 7 



3 * 



i* + 



ly 



h* - V? 



30 I 



14. 



^ 8 



ay — bz = 2 ab 
2by+ 2az = 3 b 2 — a 2 . 



3_2 



This means 



3 



3°J 



2 ^ + 3: y_ 4 2 

31- 4 ? + 22 
[4 x - 2 3; - 3 z 



17. 



r2 X - 3 y = 4 

'41 -3Z = 2 
[4 y + 2 Z = - 3, 



1.4 x +.32 ^ = 3.76 
.28X + 9.6 y=- 29.36. 



r x — y+ z = 10 
3#— 8 v + 10 z 
[Sx'+ 2 y- 32 = 




EXERCISE XI. 
Problems — Simple Equations. 

1. The sum of two numbers is 18, and the quotient of 
the less divided by the greater is ^. What are the 
numbers ? 

2. A steamer can run 20 miles an hour in still water. 
If it can run 72 miles with the current in the same time 
that it can run 48 miles against the current, what is the 
rate of the current ? 



Equations. 45 

3. A cistern has three pipes. The first pipe can fill it 
in half the time required by the second and the second 
takes two-thirds as long as the third. If the three pipes 
be open together the cistern will be filled in six hours. In 
what time can each pipe fill it ? 

4. The diameter of a driving wheel on a shaft is six 
inches greater than that of the driven wheel. While the 
driving wheel makes 64 revolutions, the driven wheel 
makes 122. What is the circumference of each ? 

5. The perimeter of a triangle is 82 in. The shortest 
side is respectively 6 in. and 19 in. less than the other two 
sides. W^hat is the length of each ? 

6. A ball is shot into the air with a certain velocity ; if 
its initial velocity had been 380 ft. per second greater, 
it would have risen \ higher in 5 seconds. What was its 
initial velocity and how high did it rise ? 

7. Jupiter is approximately 370,000,000 miles farther 
from the sun than the earth. If it takes light, traveling 
about 186,000 miles per second, 998 seconds longer to 
reach the earth from Jupiter when he is on the other side 
of the sun from the earth, what are the distances of the 
earth and Jupiter from the sun ? 

8. The distance between two points in a straight line 
is 21 miles. A train traveling on the circumference of a 
circle of which this distance is the diameter takes 13 
minutes longer to arrive at the second point than another 
train, which traveled in a straight line. If the two trains 
had exchanged tracks it would have taken the second 
train 31 minutes longer to reach the second point. Find 
the rate of each. 

9. If 1 be added to the numerator of a fraction, the 
resulting fraction will be equal to \ ; but if 1 be added to 
the denominator, the result will be \. What is the 
fraction ? 

10. A number has two digits, the tens' digit being twice 
the units' digit. If the digits be interchanged the resulting 



46 Algebra. 

number is 18 greater than the original. What is the 
number ? 

11. The sum of the lengths of two conductors is 3050 
centimeters. The resistivity (that is, the resistance per 
centimeter) of one is .0016 ohms and of the other .00972 
ohms. If each is increased 500 centimeters in length, the 
sum of their resistances is 29.66 ohms. Find their 
lengths. 

12. A boat must return to its landing in 5 hours. How 
far may it steam down stream at the rate of 15 miles per 
hour and return at the rate of 1 o miles per hour ? 

13. A and B can do a piece of work in 20 days, A and 
C in 30 days, and B and C in 40 days. How long would 
it take each alone ? 

14. A man invested $2610 in 5% and 6°/ bonds, pay- 
ing 95 for the former and 105 for the latter. His income 
from both together was $144. How much did he invest 
in each ? 

15. A train 300 ft. long passes a train 360 ft. long in 
56 J seconds when both run in the same direction ; but if 
they run in opposite directions they pass in 6\ seconds. 
What are the rates of each in miles per hour ? 

16. A mass of tin and lead, weighing 240 lbs., loses 
28 lbs. weighed in water. If 74 lbs. of tin lose 10 lbs., 
and 115 lbs. of lead lose 10 lbs. in water, how many lbs. 
of each in the mass ? 

17. Find the time between 2 and 3 o'clock when the 
hands of a watch are together. 

18. The rear wheels and fore wheels of a carriage are 
respectively 16 and 14 ft. in circumference. How far has 
the carriage traveled when the fore wheels have made 5 1 
more revolutions than the rear wheels ? 



CHAPTER IV. 



GRAPHICS. 



Article 62. A point is said to have position, but the 
statement means nothing unless its position is determined 
with respect to some fixed point or lines. Given a fixed 
point, or better intersecting straight lines, and the position 
of any point can be definitely determined by its distances 
and directions from them. For instance, the position of 
any building in a city whose streets are symmetrically laid 
out may be accurately stated by indicating its distances 
from two principal intersecting streets. 

Further, the relative positions of several points and their 



Y' 
Fig. 1. 



distances apart may be easily obtained if their dis- 
tances from two intersecting straight lines are known. 
For instance, the position of the point A is accurately de- 

47 



48 Algebra, 

fined if we say its distance from the line XX ' is 2 units 
upward and its distance from YY' is 3 units to the left. 

The distance AB to YY' is called the abscissa of A and 
the distance AC is called its ordinate for convenience. 
The abscissa and ordinate together are known as the co- 
ordinate of the point. 

Art. 63. To avoid the cumbrous expressions, " to the 
right," "to the left," etc., it is agreed that distance mea- 
sured to the right of YY' shall be called plus, to the left, 
negative ; and that distance measured upward from XX' 
shall be called positive, and downward, negative. 

Art. 64. The reference lines XX' and YY' usually 
intersect at right angles for simplicity. They clearly 
divide space into four quadrants numbered 1, 2, 3, 4, 
around to the left from X to X again. Thus XOY is 
quadrant t, YOX' is 2, X'OY' is 3, Y'OX is 4. The ab- 
scissas are universally designated by x and the ordinates 
by y. 

The point A in the last illustration is # = -3. y = 2 or 
more briefly (— 3, 2), x always being written first. 

Art. 65. The values of x and y remaining 3 and 2 
respectively, a point in the first quadrant would be 
(3, 2), in the second, (- 3, 2), in the third (-3,-2), 
and in the fourth (3, — 2), represented in the figure as B, 
A, C, andZ), respectively. 

Art. 66. Since a line may be regarded as made up of 
points, if the position of every point in the line with respect 
to the axes XX' and YY' is known, the position of the 
entire line is known. 

Art. 67. Whenever the relation between the abscissa 
and ordinate of every point on a line is the same, the ex- 
pression of this relation in the form of an equation is said 
to give the equation of the line. For example, if the ordi- 
nate is always 4 times the abscissa for every point on a 
line, y = 4.x is called the equation of the line. 

Again, if 3 times the abscissa is equal to 5 times 



GrapJiics. 49 

the ordinate plus 2, for every point on a line, then 
$x = 5 ; y + 2 is the line's equation. 

Art. 68. Clearly since an equation represents the 
relation between the abscissa x and the ordinate y for 
every point on a line, if either coordinate is known for 
any point on the line, the other one may be found by sub- 
stituting the known one in the equation and solving it for 
the unknown. 

For example, let 23/ = yx — 1 be the equation for a line, 
and a point is known to have the abscissa x= 2. To 
find its ordinate substitute x = 2 in the equation ; 2y = 7 
(2) — 1 = 14 — 1 = 13 ; y = 6Jk Therefore the ordinate 
corresponding to the abscissa x = 2 is 6\. 

Art. 69. Further, if the equation is given, the whole 
line may be reproduced by locating its points. If x for 
example be given a series of values from o to 10 inclusive, 
by substituting these values in the equation, the corres- 
ponding values of y are found, and 11 points are thus 
located on the desired line. If more points are needed 
the range of values for x may be indefinitely extended, 
and if these points are joined we have the line. For 
example, let the equation of a line be x 2 + y 2 = 9, to re- 
produce the curve represented. For convenience in cal- 
culating solve for y ; 



± v 9 - *a. 



Then give x a series of values to locate points on this 
line. 



lix = 


= y = 


■■±^9- 


= 


±3- 








lix = 


-1 y = 


±^9- 


1 = 


+ v^8 = 


± 


2.83. 




Ux = 


-2 y = 


±^9- 


4 = 


±^5 = 


± 


2.24. 




Iix = 


-3 y = 


■± V 9- 


9 = 


±v = 


0. 






lix = 


4 y = 


± v 9~ 


16 = 


±'-7 


- 


an imaginary, 



50 



Algebra. 



The last value for y shows that the point whose abscissa 
is 4 is not on the curve at all ; and since any larger values 
of x would continue to give imaginary values for y, the 
curve does not extend beyond x = 3. 

Since we have given x only positive values so far, all 
our points so determined lie to the right of the YY' axis. 
To make the examination complete, let x take a series of 



negative values, thus 



±V 9 



± V 9-4 



± Vs = ± 2.83. 
±V S = ± 2.24. 

^0 = 0. 



If x= — 1 y 
If x= — 2 y 
Ifx=-3^=± v 9~ 9 

The similarity of these results shows that the curve is 
symmetrical with respect to the axes, that it is 'alike on 
both sides of the axes. 

If now these points are located with respect to the axes 
XX' and YY' and are joined the result is an approxima- 
tion to the curve ; it is only an approximation because the 
points are few and not close enough together. 

A 




The result is shown in Fig. 2, using \ in. as a unit for 
scale. The points are (o, + 3), (o, — 3), [being A and A' 



Graphics. 5 r 

in the figure] (1/8), (i,~ ^S) [being B and B'] (2,^7), 
(2, - <5) [being C and C], (3, o) [G], (- 1,^8), (- 1, 
- V8) [£> and D'\ (- 2, Vjjj) (- 2, - </£) [E and £'] 
and (-3,0) [F\ 

Clearly if more points are needed to trace the curve 
accurately through them (as is the case here), it is neces- 
sary to take more values of x between - 3 and + 3, for 
example : 



x = 





y = 


±^9 = 


±3- 








x = 


.2 


y = 


-±^9~- 


.04 = 


= ±^8. 9 6 = 


-±2 


.99. 


x = 


•4 


y- 


± V 9~ 


.16 = 


±^8.84 = 


-±2 


.97. 


x = 


.6 


y = 


± V 9~ 


.36 = 


±^8.64 = 


-±2 


.94. 


x = 


.8 


y 


± V 9~ 


.64 = 


±^8.36 = 


-±2 


.89. 



x = 1 y= ±^g- 1 = ± ^8= ± 2.83, etc. 



Making a similar table for the corresponding negative 
values of x, the result is three times as many points on 
the curve as before, and as they are closer together the 
curve is much more readily drawn through them, and it 
will be much more accurate. 

Take another example : 9 x 2 + 16 y 2 = 144. 



Solving for y; y= ±}Vi6 — # 2 . 
Then if #=0 y= 4 J f v i6 = ±3. 



x= ±.2 y= ± j v 16 — .04= ± ^15.96= ±2.99 -I- . 

#= ±.4 y= ±}^i6^Ti6= ± ^15.84 = ±2.98 + . 
x = ±.6 y= ±f ^i6-.36 = ±^ 15.64 = ±2.96 + . 
x=±.S y= ±j v / 76^6" 4 = ± ^15^6= ±2.94. 
* = ± 1 y- ±|\/i6- 1 = ±1^15 = ±2.9, etc. 



52 



Algebra. 




Fig. 3. 



The result is indicated in Fig. 3, same scale as before. 

Art. 70. Clearly a curve can be traced thus repre- 
senting almost any form of equation. 

Suppose the equation x 3 — J x 2 + J x + 15 =yis given. 
The location of a number of points by giving x a series of 
values and calculating corresponding values of y from the 
equation, will enable us to draw through them the curve 
represented by the equation. In most cases there will be 
certain values of x which will make the value of y zero, 
such values of x will be roots of the equation x 3 — 7 x 2 + 
7 x+ 15 = o, that is these values of x identically satisfy 
this equation. 

But if y is zero for a point, the point must be on the 
x axis, for by definition the value of y is the distance from 
the x axis to the point, hence the curve must cross the 
x axis at these points where y is zero. If then none of 
the values given to x make y exactly zero, but do make y 
change from a positive value for one value of x, to a 
negative value for the next, or vice versa, it must pass 
through zero to change from one sign to the other, and 
hence the curve must cross the x axis. 

As an illustration, take the equation 

x 3 - 5 x 2 + x + 1 1 = y. 



Graphics. 5 3 

As before, make a table of values of x and y, and locate 
the points as follows : 



li x = y = 11 


If x = 3 v = -4 


If x = .5 y = 10.375 


If *- 3.5 y - - 3.875 


If x = 1 7 = 8 


If x = 4 ;y = — 1 


If x = 1.5 y = 4-625 


If «- 4.5 v = 5.375 


If x = 2 v = 1 


If ^ = — 1 7 = 4 


If x = 2.5 y = 2.125 


If x = 1.5 y = 5.125 



The curve connecting these points crosses the # axis at 
three points; one between 2 and 2.5, one between 4 and 
4.5, and one between — 1 and — 1.5. Hence the three 
roots of the equation x 3 — 5 ^ 2 + x + 1 1 = o are between 
2 and 2.5 ; between 4 and 4.5, and between - 1 and 

- i-5- 

If the values of x in the above table had been taken 
closer together, the points of crossing would have been 
more accurately known. 



EXERCISE 


XII. 


Graphs. 




Construct the graph of : 






1. 2 x - 5 = y. 


8. 


x 2 = 4y. 


2. I (* - 3) - y- 


9. 


f = J%X. 


3. 3 x - sy = 7- 


10. 


x 2 + y 2 = 25. 


4. 5 — 2 X = 2 y. 


11. 


x 3 - 3 x 2 + 1 = y. 


5 - }(4 + 5*) = \y- 


12. 


^x + -%y — 0. 


6. gx 2 + 16 y 2 = 144. 


13. 


r - 4 (4 - *)• 


7. x 2 - *y 2 - 4- 


14. 


^ = 9 (9 - y)- 



15. State the equation for the line § of whose abscissas 
minus 1 equals | of its ordinates. 

16. State the equation for the line, the sum of the 
squares of the abscissas and ordinates of its points being 
equal to 16. 



CHAPTER V. 
THE BINOMIAL THEOREM. 

Article 71. Let us consider the product (a + b) 
(a + b) (a + b) or (a + b) 3 = a 3 + 3a 2 b + sab 2 + b 3 . 

A very casual examination will show that the first term 
of this product, which is the cube of (a + b), is the first 
letter raised to the power 3, and that the exponent of this 
first letter decreises by one iu each successive term to the 
right until it disappears in the last term ; also that the 
second letter appears first in the second term and increases 
by one in each term to the right until it reaches 3 in the 
last term. 

Also the coefficient of the first term is 1, and of the 
second term is the same as the exponent of the binomial, 
that is, 3. The coefficient of the third term may be gotten 
by multiplying the coefficient of the preceding term (the 
second) by the exponent of the first letter and dividing 

the product by the number of this term ; thus, ^ — 3. 

2 

The coefficient of the fourth term (which in this case is 
the last) may be gotten in the same way from the third ; 

thus, 3JLi = !. 

3 

Now let us see if these rules hold for the fourth power 
of (a + b), 

(a + b) (a + b) (a + b) (a + b) = (a + b) 4 
= a 4 + 4 a 3 b + 6a 2 b 2 + \atf + b\ 

54 



The Binomial Theorem, 55 

Clearly the rules for exponents apply ; for the exponents 
of a begin with 4 and decrease by one to the right, and 
the exponents of b begin with one in the second term 
and increase by one, to 4 in the last term. 

Also the coefficient of the second term is 4, the exponent 
of the binomial ; the coefficient of the third term is gotten 
from the second by the rule already indicated, namely, 



4x3 



6. 



6x2 
For the fourth term also, = 4 ; and for the last 

3 
term, 1J<_i = I# 



And so an examination of higher powers will in every 
case verify these rules, which are called empirical rules, 
because they are derived by inspection and verified by 
trial, although they have a good mathematical foundation. 
These rules constitute what is called the Binomial Theorem. 

Example. Develope the expression (x + y) 7 by the 
binomial theorem. 



(x + y) 7 = x 7 + 7 x ( - ') y + 2JU> x 7 - 2 f + 2JL ^ 

2 3 

X " ' y + 35 X 4 ^ ? " ' y4 + 35 X 3 v 7 - ' y; + 



6 
x y Q + 7 X T y 7 or (x + y) 7 = x 7 +7 x*y + 21 .v 5 v 2 

+ 35 ^ 4 y 3 + 35 ^ 3 y 4 + 2 r - t2 y 5 + 7 xy* + y 7 - 



It is to be also observed that the number of terms is 
always one greater than the power to which the binomial 
is raised ; for example, in the above problem there are 
8 terms = 7 + 1, etc. Also the coefficients of the terms 



56 Algebra. 

equally distant from the first and last terms are the same, 
as may be readily seen by writing the series of coefficients 
in the last problem as follows : 



I I r— * v I I 

7> 21, 35> 35> 21, 7: 



A simple examination of the product of (a — b) (a — b) 
(a - b), (a - b) (a - b) (a - b) (a - b) etc., will show 
that the terms are alternately plus and minus, the first 
term always being plus, and that the developement in 
other respects is exactly like that for (a + b). 

Art. 72. These rules may be applied to trinomials by 
grouping two of the terms together, thus, [(a + b) + c] 4 
= (a + 6) 4 + 4 (a + bY c+6 (a + b) 2 c 2 + 4 (a + b) c 3 + c 4 ; 
then (a + b) 3 and (a + b) 2 may be developed separately 
by the theorem and the results substituted. Likewise 
(a + b + c + d) 5 could be grouped thus, [(a + b) + (c + d)f, 
and so on by like process for any polynomial. 

Art. 73. It is to be observed that these rules for the 
expansion of a binomial do not require that the terms 
of the binomial should be single letters ; for example, 
(2x + 2>yY can b e expanded in exactly the same way 
as (a + b) 4 , by treating the 2x as a and 3V as b\ thus 
(2x + 2>yY = ( 2 #) 4 + 4 ( 2 ^) 3 (3y) + 6 (2x) 2 (3y) 2 + 4 {2%) 

(3y) 3 + (3y) 4 

This last expression can then be simplified by perform- 
ing the indicated operations as follows : 

{2xy - ( 2 ) 4 (x 4 ) - 16 x 4 ( 3 y) 2 - (3) 2 (f) = 9 f 

(2XY = (2)3 ( X ^) = 8X 3 ( 3 yY = (3) 3 (f) = 21 f 

(2x) 2 - (2) 2 (x 2 ) = 4 * 2 (3^) 4 = (3) 4 (/) - 81 / " 

Making these substitutions, the first expression becomes 
(2x + 3y) i== 16 x 4 + 96 x 3 y + 216 x 2 y 2 + 2i6x^ 3 + 8i y 4 . 



The Binomial Theorem. 57 

Clearly in the binomial [a + b), a or b can be made to 
represent any expressions whatever, and hence any polyno- 
mial, however complex, may be expanded by the theorem. 
The binomial theorem may be employed to abbreviate 
the process of squaring numbers, especially those of two 
digits, as 47, say. The process is as follows: 

(47 2 = (40 + 7) 2 = (40) 2 + 2 x 40 x 7 + 7 2 by the theorem. 

1600 
This may be collected thus, 560. Since the first digit 

is of necessity in ten's place 4q its square will be 

2209 
always so many hundredths, and hence there will be always 
two zeros with the square of the first digit; twice the 
product of the two digits will always have one zero with 
it as the second term of the development. The addition 
can be simplified then by neglecting the zeros and keeping 
the figures in their proper position in the sum by writing 
each succeeding term so that one figure projects tb the 
right, thus, 





( 47 ) 2 = 16 = 4 2 




56 = 2 x 7 X 4 




49 = 7 2 




2209 = (47)2 




EXERCISE XIII. 




Expand by the Binomial Theorem. 


1. 


(a + by. 5. ( 3 c- 5 dy. 


2. 


(2 * + 3 y) 5 . 6. (2 - x) 7 . 


3. 


(x - 2Z) 3 . 7. (y - I) 8 . 


4. 


(- + A' - 8. (xy - z) 3 . 



58 



Algebra. 



9. 


(3-< 2 ) 5 . 


19. 


(f + y- i) 3 . 


10. 


(i - x 2 ) 6 . 


20. 


(2 + X + X 2 ) 4 . 


11. 


(x + |) 5 . 


21. 


(\/^+ j)5. 


12. 


(m - § w) 4 . 


22. 


(XV — 2) 4 . 


13. 


(x 3 — y 2 z) 3 . 


23. 


(* - I) 6 . 


14. 


(r~i + ,?~3) 5 


24. 


(a -x) 7 . 


15. 


(2x* + y-?y. 


25. 


(i + x — y —z) 3 . 


16. 


( 3 V^ _ 2 \^)4. 


26. 


(37 - i) 5 - 


17. 


(a§+ 5 ) 5 » 


27. 


(* + *«)<• 


18. 


f.+.^Y 


28. 


(* + AY- 



X] 



INVOLUTION AND EVOLUTION, 

Art. 74. The process of multiplying any quantity by 
itself any number of times is called raising it to the 
power indicated by the number of times it is used as a 
factor. For example a X a X a = a 3 is the third power 
or cube of a. Power in algebra differs from power in 
arithmetic chiefly in the matter of sign. From the rule 
for signs in multiplication, evidently all even powers of 
either positive or negative quantities will be positive, 
while odd powers of negative quantities are negative. 
Raising to powers is called Involution. 

Art. 75. To raise a monomial to any power, raise the 
numerical part arithmetically to the required power, and 
multiply the exponents of the literal factors by the power, 
as directed in the rule for exponents. 

Example. 

(5a 2 k 3 ) 3 =(5) 3 fl 2 x 3 & 1 x 3 c 3 X3 =I25fl 6^9 ( _ 3I -i ;y 2 z) 5 

- (- 3) 5 x- lx5 v 2X5 2; 1 x 5 = - 24 3 x- 5 v 10 z 5 . 

Art. 76. To raise a polynomial to any power, apply 
the binomial theorem as indicated in Article 72. 



The Binomial Jlieorem. 59 

EXTRACTION OF ROOTS. 

Art. 77. The root of a quantity is one of its equal 
factors. Evidently the finding of any root is the exact 
reverse of raising it to a power. 

Hence, to extract a root of a monomial, extract arith- 
metically the required root oj the numerical factor, and 
divide the exponents of the literal factors by the index of the 
root. 

Roots are indicated as in arithmetic by the radical sign, 
V and the figure indicating the root called its index, 
above the radical sign to the left. Thus, ^65 a 2 b indi- 
cates the fifth root of 65 a 2 b. This is often expressed 
as a fractional power, thus (65 a 2 b) i Extraction of 
roots is known as Evolution. 

To Extract the Square Root of a Polynomial. 

Art. 78. Since by making use of parentheses, and thus 
grouping terms, any quantity may be put into the form 
of a binomial, its square may be regarded as the square 
of a binomial, and the simple binomial form (a + b) 2 
= a 2 + 2 ab + b 2 may serve as a model. In this expres- 
sion, a and b are to stand for any two quantities, mono- 
mial or polynomial. For example, 

(2 x 2 + 3 c y + m 3 + n) 2 = [(2 x 2 + 3 c y) + (m 3 + n)\ 2 
= (2 x 2 + 3 c y) 2 + 2 (2.V 2 + 3 c y ) (m 3 + n) + (m 3 + n) 2 . 

The expression (a + b) 2 = a 2 + 2a b + b 2 may be put in 
the form (a + b) 2 = a 2 + b (2a + b). Since as we have 
shown any quantity may be expressed in this general form, 
the square root of the first term of such quantity is the 
first term of its complete square root, for V a 2 = a in 
the form (a + b) 2 = a 2 + 2 a b + b 2 — a 2 + b (2a + b). 
If now a 2 be subtracted from the square of the binomial, 



6o Algebra. 

the remainder 2 a b + & 2 = b (2 a + b) will contain the 
rest of the root. Since the first term of this remainder 
is made up of 2 a and b, if we divide it by twice the first 
term of the root already found (a), the quotient will be b, 
the second term of the root. 

The form of the remainder, b (2 a + b), shows that to 
make the process complete, b must be added to 2 a, and 
the sum.multiplied by b. 

If this when subtracted still does not exhaust the origi- 
nal quantity, the same process repeated using the two 
terms of the root already found, a + b, as a single term, 
will add another term to the root. This would show that 
the original expression was in the form (a + b + c) 2 
instead of (a + b) 2 , thus, 

[(a + b) + c] 2 - (a + b) 2 + 2 (a 4- b) c + c 2 * 

Now (a -f b) 2 has already been subtracted from the 
above expression in the process just described, hence 
the remainder consists of an expression of the form 

2 (a + b)c + c 2 = c [2 (a + b) + c], 

which is exactly like the first remainder, if a ■+ b is regarded 
as a single quantity, indicated by the parenthesis. 

This process may be repeated until there is no re- 
mainder, or one too small to further contain the terms of 
the root already found. In the latter case there is no 
exact root. 

Art. 79. All this maybe put into a rule as follows: 

To extract the square root of a polynomial, extract the 
square root 0} the first term, this will be the first term of 
the root. Double this 'first term 0} the root, and as a trial 
divisor, divide the product into the remainder left after drop- 
ping the first term of the polynomial ; the quotient is the 
second term cf the root. Complete the trial divisor by add- 






The Binomial Theorem, 



61 



ing this second term of the root, and multiply this sum by 
the second term of the root. If, after subtracting this pro- 
duct, there is still a remainder, double all of the root already 
found and divide the first term of this product into the first 
term of the remainder to get the third term of the root. 
Complete the divisor as before and multiply the completed 
divisor by the third term of the root. Continue this process 
until there is no remainder. The polynomial should be 
arranged according to ascending {or descending) powers of 
one letter. 

Example. Extract the square root of 

25 x 4 + 2ox 2 y— 4-yz 2 + 4 y 2 + z 4 — 10 x 2 z 2 

25 x 4 + 20 x 2 y + 4 y 2 — 10 x 2 z 2 — 4 yz 2 + z 4 k x 2 + 2 y — z 2 

25 x 4 



Trial divisor : 10 a" 2 

+ 2 y 



: ox 2 y -f 4 y J — 1 o„v 2 z 2 — 4 yz 2 + z 4 



2Qx 2 y+ 4y 2 



Complete div. io3; 2 + 23; 
(101 2 + 2 y) 2 y = 
2 (5 x 2 + 2 y) = 10 x 2 + 4 y (trial div.) - 10 a?z 3 — 4 yz 1 + z 4 

-_z 2 

iox 2 + 4y — z 2 (complete) 
(iox 2 + 4y - z 2 ) (- z 2 ) = 



I O X"Z" 



\yv 



This expression was of the form {a + b + r) 2 , 

for [(5 „r 2 -f 2 v) - z 2 ] 2 = (5 x 2 + 2 y) 2 + 2 (5 jc ? + 2 y) 
( - z 2 ) + ( - z 2 ) 2 =25 x 4 +20 afy - 4 jz 2 + 4y 2 + z 4 - 1 o x 2 z 2 . 



Extraction of Cube Root of Polynomials. 

Art. 80. A similar consideration of the cube of the 
binomial (a + b), (a + b) 3 = a 3 + 3 a 2 b + 3 tf6 2 + fr 3 =j 3 
+ 6 (3 a 2 + 3 tf& + &'-)> will make the following rule clear: 

To extract the cube root of a polynomial, arrange, for 
convenience, the polynomial according to the ascending or 
descending power of one letter. Extract the cube root of 



62 



Algebra. 



the first term ; this will be the first term of the root. Mul- 
tiply the square of this first term of the root by 3 (3 a 2 in 
the above formula) for a trial divisor ; this product divided 
into the first term of the remainder left by dropping the first 
term of the polynomial, will give the second term of the 
root (3 a 2 b -5- 3 a 2 = b), complete the divisor by adding to 
the trial divisor three times the product of the first and 
second terms of the root and the square of the second term 
of the root (3 a 2 + 3 ab + b 2 ) ; multiply this sum by the 
second term of the root ( [3 a 2 + 3 ab + b 2 ] X b) and sub- 
tract from the remainder of the polynomial, jf there is 
still a remainder large enough to contain the trial divisor, 
repeat the process, taking the two terms of the root already 
found together as a single term. 



Example. Extract the cube root of 

63 x 4 + 27 x 6 + 21 x 2 - 44 x 3 - 54 x 5 - 6 x + 1. 



Arranging : ?. "Jj 

2 7 x% — 54 x 5 + 63 oc A — 44 x 3 + 2 ix 2 — 6 x + 1 k x 2 — 2 x + 
2 7 x Q 

Trial Divisor : 

(3X 2 ) 2 3 = 27 X 4 
(3X 2 ) ( - 2x)3 = — i8x 3 

(— 2X) 2 = +4X 2 



27 4 - 18 X 
Complete. 



3 + 4x 2 



54x 5 + 63X 4 — 44X 3 + 2 ix 2 — 6x+ 1 
54x 5 + 36x 4 -8x 3 

= (2 7 4 - i8x 3 + 4x 2 ) (-2x) 



Trial Divisor : 
(3 X 2 — 2 x) 2 3 = 27 x 4 — 36 x 3 
+ 12 x 2 
( 3X 2 - 2x) ( + 1) (3) = 9X 2 — 6x 

(I) 2 - ~ + 1 



2 7# 4 - 36 x 3 + 21 x 2 — 6 x + 1 
Complete. 



27X 4 — 36x 3 + 2 ix 2 — 6x+ 1 



27X 4 — 36x 3 + 2ix 2 — 6x+ 1 
= (27X 4 — 36x 3 +2ix 2 
- 6x + 1) (+ 1) 



The Binomial Theorem. 63 

Art. 81. Evidently similar rules for higher powers 
may be readily stated by referring to the corresponding 
developments of (a + b), but they become rather compli- 
cated and are rarely needed. 

The fourth power being the square of the square, the 
fourth root may be gotten by extracting the square root 
twice. Likewise, the sixth root by extracting successively 
the square and the cube root, or vice versa, etc. 

If it be remembered that numbers may be expressed as 
binomials (as 67 = 60 + 7 or 139 = 130 + 9, etc.) the 
arithmetical application of these rules will be readily 
understood. 



LXtl 


EXERCISE XIV. 

•act the following roots : 


1. 


^a 2 b 2 (a + b) 4 . 5. V6 4 *y 8 z 30 . 


2. 


V25 bx 8 y l2 a 20 . 6. $/ - 729 (a + b) 3 c 6 . 


3. 


V - 32 ;;/ 10 /rV 15 . 7. vVy% u . 







y/2-«V. 



8. v / I 6 ( r + s) (r + s) 3 . 



Extract the square root of the following : 

9. 13 x 2 y 2 — 12 x 3 y + 4 x 4 — 6 xy 3 + y A . 

10. 4 m 4 - 12 m 2 n 2 + 9 n A + 16 m 2 p 2 - 24 n 2 p 2 + 16 p 4 . 

11. 9 — 24 y — 68 y 2 + 1 12 y 3 + 196 y 4 . 

12. 4 + 9 6 2 — 20 a + 25 a 2 + 30 #6 — 12 6. 

13 x l + Z 6 + -! 2 _ A " 2 v ' 3 j_ _J Z 3 . 

4 9' 4 3 16 6 



64 Algebra. 

14. HOL - ±Ht +4 m 2 n 2 + 6m - 12 n 3 + 2J»V 



15. / - 6 /s + 15 z 2 / - 20 z 3 / + 15 z 4 y 2 -6 z 5 y + z 6 . 

16. x 2 + 4 x 3 + 4 ^ 2 + 4 x + 4 + - . 

Extract the cube root of the following : 

17. a e + 3 a 5 + 6 a 4 + 7 a 3 + 6 a 2 + 6 a + 1. 

18. jc 3 — 6 x 2 ^ 4- 1 2 x;y 2 — 8 y 3 — 3^2 + 3X2 2 + 12 ^z 
— 12 y 2 z — 6 yz 2 - z 3 . 

19. 8 ;y 6nt - 36 y 5m + 66 y 4m - 63 y 3m + ^^ y 2m - 9 /" + 1. 

20. x 3 - x 2 y + ^ - £- . 

3 27 

21. I2a 2__il5 _ 54a _ 59 + _£35 + 8a 3 + li. 

22. 1 - 6 ;y + 9 y 2 + 9 y 3 - 9 / - 6 / - /. 

23. ^ + — 2_. + 12 x - 7 - 12 x 2 + 8x 3 . 

x 8 x 3 4r 

24. 27 m 6 ;z 6 + 54 m 5 n 5 + 9 m% 4 — 28 m 3 n 3 — 3 m 2 w 2 
4-6 ww— 1. 



CHAPTER VI. 

SURDS. 

Article. 82. It is understood that the expression x 5 
means that x is to be taken five times as a factor ; that is, 

/y* sy* -V* "V* / V* /y*« 

It is found convenient to represent in somewhat similar 
symbols other operations involving factors. These sym- 
bols have been suggested by the simpler operations with 
integral exponents, thus : 

Each of the successive expressions a 6 , a 5 , a 4 , a 3 , a 2 , a 1 , 
can be derived from the next preceeding by dividing by 
a, that is, a 5 = a G -j- a : a 4 = a 5 + a, etc., so that each 
division reduces the exponent by unity. If we continue 
this process in above series, we get a 6 , a 5 , a 4 , a 3 , a 2 , a 1 , a , 
a~ 1 , a~ 2 , a -3 , a~ 4 , etc. 

By the primary laws of division 

a 1 -h a 1 = 1 = a by above series. 

a -+ a = — =■ a~ l by above series. 
a 

1 

a -1 + a = = — = — = a~ 2 by above series. 

a a a 2 

Hence, we may adopt the symbols : 

a- 1 for — , a~ 2 for — , a~ 3 for — , etc. 
a a 2 a 3 

Also a = 1. 

o- ,<7 -1 1 , , a~ 2 1 

Since a~ x = = - and a~ 2 = = -, etc. 

1 a 1 a 2 

65 



66 Algebra. 

We may state the rule thus : 

Any quantity may be moved from denominator to 
numerator of a fraction, or vice versa, by changing 
the sign of its exponent, or: 

a quantity with a negative exponent is the recip- 
rocal of the same quantity with the exponent made 
positive. 

FRACTIONAL EXPONENTS. 

Art. 83. Let us consider a series of expressions like 
the following : 

a 16 , a 8 , a 4 , a 2 , a 1 . 

Each one obtained from the preceeding, by dividing its 
exponent by 2, is called its square root. If this process 
of division is continued, we get, 

a 16 , a 8 , a 4 , a 2 , a 1 , a*, a*, a£, etc. 

and by analogy, cfi may be called the square root of a ; at, 
the square root of a% or the fourth root of a, etc. 

Likewise, the series a 27 , a 9 , a 3 , a 1 , ai, a?, a&, etc., indi- 
cates a like relation with third, ninth, twenty-seventh root, 

and in general, the symbol a H means an nth root of a. 

RADICAL SIGNS. 

Art. 84. There is another symbolism which arises 
from the older method of indicating square root by pre- 
fixing the Latin word radix, meaning root, to the quantity 
involved ; for instance, radix 2 meant, in modern phrase- 
ology, the square root of 2, etc. 

Eventually radix was abbreviated to r, and finally the 
r was extended to cover the quantity whose root was 
required ; thus, ^2 • 

Hence a Hs equivalent to ^a. 



Surds. 67 

To complete this symbolism, a small index is super- 
posed upon this radical sign, to indicate other roots, thus : 

a* = \/a,a = -tya, and in general a 7t = ^/a. 

Art. 85. Any root thus indicated, except even roots 
of negative numbers (discussed later), is called a surd. 

Art. 86. Surds are said to be similar, when they 
have the same quantity under the radical sign, when in 
there simplest form. 

av/ 3> 3^3> 2 ^3> etc., are similar. 

Art. 87. When an expression is wholly under the 
radical sign it is said to be a pure surd; otherwise, a 
mixed surd. 

^h ^9> y/ ^ I 3i etC- ' are P ure sur< ^ s - 
3V2, 5^7, 4^3, etc., are mixed surds. 



To Simplify Surds. 
Art. 88. Example, simplify 



^27a 3 b 2 c'° = ^ (ga 2 b 2 c 4 ) ^ ac ) = ^abc 2 ^ac, 

since the square root of 9 a 2 b 2 c 4 is 3 abc~. 

Again, ^216 = ^36 x 6 =6^, 

or ^81 = ^27 X3 = 3 VJ. Then, 

Rule: Separate the expression under the radical 
into two factors one of which is a perfect power, and 
the largest factor of this kind. extract the root 
(indicated) of this factor, placing it outside of the 
radical sign, leaving the other factor under the 
radical. Multiply the root that has been removed 
by any factor already outside of the radical sign. 



68 Algebra. 

Example. Simplify, 3^98 

3 Vg8" - 3 ^49 x 2 = 3 (7) V7 = 2 1 ^2. 
Again, ab^a 3 b 5 x 2 = ab^(a 2 b 4 x 2 ) (ab) = a 2 b 3 x ^ab. 

Order of a Surd. 

Art. 89. The index of the radical indicates the order 
of the surd. For instance, v 5 is a surd of 2d order, or a 
quadratic surd ; V5 is a surd of 3d order, or a cubic 
surd ; ^5 is a surd of 4th order, or a bi-quadratic, etc. 

Reduction of Mixed to Entire Surds. 

Art. 90. The reduction of mixed surds to entire surds 
is the exact reverse of the simplification of surds. 

Example. 2^3" - x/ (2) 2 X3 = "^4X3 = ^12 



Again, 3 V7 = V(3) 3 X7= ^2jxy= V 189, etc. 
Rule: Raise the entire expression outside the 

RADICAL SIGN, TO THE POWER INDICATED BY THE RADICAL 
INDEX, MULTIPLY THE RESULT BY THE QUANTITY ALREADY 
UNDER THE SIGN, AND WRITE THE PRODUCT UNDER THE 
SIGN. 

Addition and Subtraction of Surds. 

Art. 91. Manifestly radicals involving different 
quantities cannot be added or subtracted. For instance, 
3 V2 and 2 ^5 cannot be added or subtracted, except by 
indication, as 3^2 ^ 2^5, for the square root of 2 is 
very different from the square root of 5. It would be as 
possible to add 36 and 2a. 



Surds. 69 

But 3 v 2 , 2V2, 5^2, etc., can easily be added as 3a, 
2a, and 5a ; thus, 3 ^2 + 2 V2 + 5 ^7 =10 ^2". 

Nor can surds of different orders be combined for the 
same reason. For instance, 3^ ^2,4 \l 2. cannot be 
added, because each is entirely distinct, like a, b, c, etc. 

Multiplication and Division of Surds. 

Art. 92. 3^2 and 2^2 are as distinct entities as a 
and 6, for the square root of 2 is plainly as different from 
the third root of 2 as if it were an entirely unlike quantity 
under the radical sign. 

If it were required to multiply (a + b) 2 by (a + b), it 
would be incorrect to multiply (a + b) with the exponent 
2, by (a + b) with exponent 1, thus, (a + b) 2 X (a +b) 
= (a 2 + 2ab + 6 2 ) 2 , because (a + &) 2 is of a quite different 
order from (a + 6) 1 . It is then necessary to reduce both 
to the same exponent, thus, (a + b) 2 = (a 3 + 20b + 6 2 ) 1 x 
(a + b) 1 = [(a 2 + 2ab + 6 2 ) (a + ft)] 1 - (a 3 + 3 a 2 b + 3 ab 2 
+ P)\ or (a + 6) 4 X (a + b) 2 = (a?+2ab + 6 2 ) 2 X (a + 6 2 ). 2 
Likewise, if 3^2 is to be multiplied by 2 V2, these surds 
must be reduced to the same order. 

The least common order to which each of these surds 
can be reduced is clearly the 6th ; thus : 

3 ^2 = 3 i/JSp-iV* 

2 %/ 2 = 2 V(2) 2 = 2 V4. 

Then 3^8 x 2 ^4 = (3 x 2) ^4x8 = 6^32. 

Rule: Reduce the surds to the same order. Mul- 
tiply THE COEFFICIENTS OF THE SURDS TOGETHER, FOR 
THE COEFFICIENT OF THE PRODUCT, AND THE QUANTITIES 
UNDER THE RADICAL SIGN FOR THE SURD PART OF THE 
PRODUCT. 



?o Algebra. 

Division of Surds. 

Art. 93. Division likewise requires the reduction of 
the surds to the same order. Surds of different order 
can no more be multiplied or divided by one another than 
can bushels be multiplied or divided by feet. Hence, 

Rule: Reduce the surds to the same order. Di- 
vide THE COEFFICIENTS AND THE QUANTITIES UNDER THE 
RADICAL SIGN SEPARATELY AND EXPRESS THE QUOTIENT AS 
THE PRODUCT OF THESE TWO PARTIAL QUOTIENTS. 

Example. Divide 3 a *s/a 2 b by 2 ab V 'ab 2 . 
SaV^b = sa^Ja^by = 3 a '-vVF 



2 ab \lab 2 = 2 ab V (ab 2 ) 3 = 2 ab V a 3 b 6 

30 Var ^ 20b va 3 b Q Ti2/ — 3 Ti ~ ~L\ T; 

2 ab \Ja 6 b b 2 y b 2 



or $■ b- 1 tya 5 &~ 2 . 

2 

The order to which the surds are reduced should be 
the L.C.M. of the indices of the original surds. 

Comparison of Surds. 

Art. 94. If it is necessary to compare any quantities, 
they must be expressed in the same unit. Likewise, if 
surds are to be compared, they must be reduced to the 
same order. Hence, 

Reduce the expressions to complete surds. Re- 
duce THESE COMPLETE SURDS TO THE SAME ORDER, AND 
THE SURD HAVING THE GREATEST QUANTITY UNDER THE 
SIGN IS THE GREATEST. 



Surds. J I 

For example, compare 2 V3, 3^2, and f V4. 



2^/3 = ^8x3= ^2^ 

3 V7 = v 9 x 2 = ^Ts 

f ^4 = 4 /62 5 X 4 = 4 / 62 5 = ^156.25" 



4 /62 5 x 4= 4 /62 5 



12 is the L.C.M. of the indices. 

^78 = ^(i8) 6 = ^34012224 

-y 150.25= l ^'(i 5 6.2 5 ) 3 = ^38,41 1.389385. 

Hence, ^18 is the greatest. 

Rationalizing Denominators Containing Surds. 

Art. 95. It is usually undesirable to have surds in the 
denominators of fractions, owing to the difficulty of esti- 
mating values in such forms. 

The principle derived from factoring, that the product 
of the sum and difference of two quantities equals the 
difference of their squares, enables us to remove these 
surds. 

For example : To rationalize the denominator of the 

fraction -7= 

V 7 -2 

Since the multiplication of both terms of a fraction by 
the same quantity does not change its value, we can choose 
any multiplier we please. If we choose the quantity 

^7 + 2, we will have the sum of the quantities ^7 and 2, 
whose difference is the denominator of the fraction, and 
since the product of this sum and difference gives the 
difference of their squares, the radicals will disappear. 
Thus : 

6 "^7 + 2 6^7 + 12 6^ + 12 ,- 

-7= X —r= = — 7= — = '— - =2^7 + 4. 

^7-2 ^7 + 2 (V 7 )2_ (2) 2 7 _ 4 



> Algebra. 

Again, to rationalize the denominator of 







2+v 3 






v7- 


- V 3+ V 5 


This fraction 


may 


be written thus 








i 


i + v 3 






(VI- 


- V 3) + 


^5 






(^2- 


-^)- 


^S 






(^2 


- V 3)- 


^7 


Multiply by 










2+ V 3 


N< 


/(V^. 


- V 3)~ 


^ 


{ v 2 -y 3 ) + 


^ 


\v 2 . 


- V 3)- 


^5 



_ 2 v / 2 _ 2 v / 3_ 2 v / 5+ \/6- Vi 5 -3 
( x/ 2 _V3 )2 _ ( V^ )2 

_ 2V / 2-2V / 3-2^5- V I 5+ ^6-3 
2^6 

Multiply again by — — ; 

2^-2^3-2 v^ + ^6- ^"15- 3 ^6 ' 

X 



^6 V6 



2^12-2^18-2^30 + 6- ^90-3^ 
12 

4 V^ _ 6 V 2 -2^30 + 6-3^10-3^6 



I VT _ I V7 - 1 V 30+ I -I V IO _ I v^. 
3 2 6 24 4 

This last expression may be readily computed. 



Surds. 73 

Art. 96. A quadratic surd in its simplest form cannot 
equal the sum of a rational quantity and a surd. 

For if it were possible let v x = y + Vz. 
Squaring both sides x = y 2 + 2y^z + z\ 

transposing, x - y 2 -z = 2 y^ z\ 

that is, a surd is equal to a rational quantity, which is 

manifestly impossible. 

.-. ^x^y + v'z. (^ means wo/ equal.) 

Art. 97. The sum or difference of two dissimilar 
quadratic surds, in their simplest form, cannot equal a 
rational quantity or be expressed as a single surd. 

If it were possible let v # ± ^ b = c 

Squaring, a ± 2^ ab + b = c 2 ; 

transposing, ± 2"^ ab = c 2 — a — b\ 

but a surd cannot equal a rational quantity, 

.-. ^7i ± ^b^c. 
if v~j ± vj } = v7. 

Squaring, a ± 2 v ab + fr = c ; 
transposing, a + & - c = ± ^2 ab, 

which is impossible, .-. ^a ±. v ' b^^ c. 

Art. 98. If the sum of a rational quantity and a surd 
equals the sum of a rational quantity and a surd, then the 
rational quantities must be equal and the surds equal. 

That is, if ^ x + m = v ' y + n, x = y and m = n. 



^4 Algebra. 

For, transposing, ^x — v ' y = n — m. 

This violates the previous article unless both sides are 
equal to zero, that is, unless x = y and n = m> 

Roots of Quadratic Surds. 
Art. 99. 
Example. Extract the square root of 9 + 4^5- 

Suppose ^9 + 4 ^5 = v^ + Vy. 
Squaring, 9 + 4^5 = x + y + 2^ xy. 

By last article x + y = 9 (1) 

and 2 ^ xy = 4 v/ 5 (2) 

or x + y = 9 (1) 

squaring (2) and dividing by 4 

xy = 20 (3) 

Equations (1) and (3) tell us that the sum of the two 
numbers x and y is 9, and their product 20. 

Evidently the numbers are 4 and 5 ; say x = 4 and 
y= 5, as it makes no difference about the order of arrange- 
ment of the numbers (y = 4 and x = 5 would serve as 
well). 

\ ^9 + 4^5= V^ + V^„ V~l+V^ =2+ y/J m 

Verify, (2 + ^) 2 = 4 + 4^5 + 5 =9 + 4 V 5~- 

Evidently a surd with a minus sign, as 4^5 — 9, would 
have the form (V# _ v^) 2 . 



Surds. 75 

ANOTHER METHOD. 

Art. ioo. Since the squares of the sum and of the 
difference of two surds differ only in the sign of the surd 
terms in these squares : 

as, {y~x + v/ y) 2 = x + y + 2 ^ xy 

(Vx - ^ y) 2 = x + y - 2^ xy y 

in the example just solved 

if (VX+ V~y)2 ==() + 4 V7, 

then (\/ x _ V y )2 = 9 _ 4 V 5> 



or V x + V y = ^ 9 + 4 \/J 



Multiply, x - y = ^(q) 2 - (4^5) 2 = Vi = 1 

as before x + y = 9. 

Add, and 2 x = 10 x = 5 

subtract, 2 y = 8 y = 4 

as before. 



EXERCISE XV. 

Simplify : 



1. ^48 x 5 b«c 8 . 3. V 4 a 3 ft _ 8 a 2£ 2 + 4 abK 

2. V a 3 _ 2 a 2 b + a6 2, 4 . ^!6a 5 x 9 . 



5. V(a 2 - 2 ax + x 2 ) 3 . 
Express as complete surds : 

V ^ + 



2x3/ + y 2 



7. $abc^a- 2 bc-\ 9. 4^4, 3^5> 5^3- 



7 6 



Algebra. 



Perform indicated operations and simplify 



Vi 



10. £^ x 7 - 



V 



48 



3 v 27 3^14 i5 V21 

11. 1^2 1 * T6 ^^V* 

12. V27 4- 2^48" + 3^108. 

13. 2^3 + 3^4"- V 5i^_ 

14. ^54 + 3 ^16 + ^43^ 

Reduce to same order : 

15. 3 ^+^ / 6and2\/i. 16. 5^^! y/7, 6 ^3. 
17. 9^4 ^ 5 ^3~ (which is the largest ?). 

Rationalize denominators : 

18. VTh, v^" i 



1 + 



V7 



19. 



20. 



21. 



3^4 2-^3 



- v, 



2 _ V 3 



3 + 4 V J_ 



1 + ^2 + 
8 

v 3 + v7 



V 3 V6 + ^2~ - ^5" 



v7_ v 3 



Extract the square root 

22. 12 - 2^35. 

23. 4 + 2>3- 



26. 18+8V5 



24. 8 - 4 x/ 3- 

25. 70 - 30 V 5 



Surds. 77 

IMAGINARIES. 

Art. ioi. Since the squares of both positive and 
negative quantities * are positive always, the square root 
of a negative quantity is something essentially different 
from the quantities heretofore considered. 

Such roots of negative quantities are called imagin- 
aries. They will arise occasionally in the solution of 
quadratic equations. 

Art. i 02. A pure imaginary is of the form 
a ^T or a v^^^T. 

A complex number is of the form a ± v/ — b. Define 
each. 

Imaginaries are added and subtracted according to the 
usual rules for surds. 



Multiplication of Imaginaries. 

Art. 103. The application of the general laws of 
multiplication must be made with some care in the pro- 
ducts of imaginaries. 

For example: (^ - a) x (V - a) does not equal 
' >/ fl 2 = ±i/ as the ordinary process would indicate, but 

(v - a) x ( v - a) = -a, which restricts the value 
of this product to one value instead of two ( + a and —a). 
It is customary and simplest to reduce all imaginary 
terms to the form Va v/ — 1 (V — a), wherein the factor 
V _ 1 always appears and the general rule, that the 
multiplying of a quadratic radical expression by itself re- 
moves the radical sign, may be applied without confusion. 

* A pure number (or its representative, a. letter) independent of 
sign is often called a scalar number or merely a scalar. 



7 8 Algebra. 

Hence : 

V~^b(- i) = - y/a£ 



= - ^ab{- i) = + ^a6. 



(+ V - a) X ( - v .ft) = + V a& . 
( - ^~^) X ( - ^ab) = ( - ^a ^^7) X ( - V£" V37) 
- (+ V^ft) (- i) = - ^~ab. 

Art. 104. The product or quotient of two complex 
quantities is in general a complex quantity. Verify this 
by examples. 

Every complex quantity can be expressed in the form 
a + b ^ — 1 evidently. 

Art. 105. Two complex quantities both consisting of 
the same terms, but united by contrary signs, are called 
conjugate complex quantities ; as, 

a + b V - 1 and a - b V - 1, 

or — x + y V — 1 and — x — y V — 1. 

Art. 106. The product of two conjugate complex 
quantities is a real quantity. For example, 

(5 + 2 v^=l) (5 - 2 vTT-i) = [ 5 2 - (2 V~) 2 ] 
= 25 -(- 4) = 25 + 4 = 29, 
or (— m+ n v — 1) {— m + n V — 1) 

= [( - m) 2 - (» V- i) 2 ] = m 2 - ( - n 2 ) 
= m 2 + n 2 , etc. 

Art. 107. Clearly the sum of two conjugate 
complex quantities is real and their difference is a 
pure imaginary. 



Surds. 79 

Example, (a + b V — i) + (a — b V — i) = 

d + JV-i+a-JV-i =2 a. 
Also, (a + 6 V - i) - (a - b V - i) 

= a + 6 V - i — a + & V— i = 2 b V — i. 

Art. io8. By reference to the similar propositions 
under surds, it will be clear that : 

If two complex quantities are equal, their real parts 
must be equal and their imaginary parts equal. Thus : 

if x + y V — i = m + n V — a , x = m 

and y V — i = n v' — i or y = n. 

If a complex quantity equals zero, both real and imagin- 
ary parts are zero. Thus, 

if a -f b V - i = o, 

a = o and 6 = o. 



EXERCISE XVI. 

Multiply : 

1. 4 + x/"^"3 by 4 - \/^~3. 

2. V3 — 2 \/ — 2 by V^ + 2 \^ 

3. 5 + 2 v'""^ by 3 - 5 \/ - 2. 



Divide and rationalize denominators : 



4. 26 + (3 + V - 4-) 

5. (3 + V^T) - (4 + 3 V^T7) 

6. 63 ^/^T6" - V~~&i. 



7. (a + V- x) + (a - V - x.) 



8. 1 - (3 - 2 V - 3). 



CHAPTER VII. 
INDETERMINATE EQUATIONS. 

Article 109. A system of equations containing a less 
number of equations than of unknown quantities, is called 
indeterminate. For instance, one equation containing two 
unknowns is an indeterminate ; two equations containing 
three unknowns are indeterminate, etc. 

Art. 1 10. As the name "indeterminate" signifies, 
such equations have no single solution. 

For instance, an equation like 3X + $y = 16 may be 
satisfied by an infinite number of values of x and y, for we 
may give either one any value we please, and by substi- 
tuting it in the equation, find the value of the other 
unknown that, with it, will satisfy the equation. 

Art. hi. We may, however, limit the number of 
solutions, by confining the values that the unknowns may 
have, to integers, and still further limit them by specifying 
that the values of the unknowns must be positive integers. 

The latter offers the only aspect of practical interest to 
these equations, because we deal in real experience only 
with positive quantities and largely with integers. 

Art. 112. As an illustration of how these restrictions 
affect the number of solutions, consider the equation above, 
3 X + Sy = *6- Without any restrictions 31 + 5y= 16 
may by satisfied by any of the following sets of values : 






X = 


y - 3i 


X = — I 


y - 3l 


X = 1 


y = 2 f 


X = — 2 


y - 4? 


X = 2 


y = 2 


x = - 3 


y = 5 


x = 3 


y- if 


x = — 4 


y - 5f 


x = 4 


y-t 


* = - 5 


y = 6 F 


x = 5 


y = i 


x = - 6 


y-6* 


x = 6 


y = -i 






and so on indefinitely. 







So 



Indeterminate Equations. 8 1 

If we restrict the values to positive quantities, all the 
negative values above will be excluded, if we restrict the 
values to positive integers, it removes all but one pair of 
values from above. 

Art. 113. It is often desirable then to determine the 
possible solutions of an indeterminate equation in terms of 
positive integers. 

It is plainly quite impracticable to make out a complete 
list of values, for they are infinite ; but it is possible to 
arrive at the result in another way, indicated in the fol- 
lowing solution. 

Example. Solve in positive integers 3 x + $y = 16. 
Divide by the coefficient of one of the unknowns, prefer- 
ably by the smaller, in this case 3. 

x • 2V 1 

It gives, x + y H = 5 + — . 

3 3 

rp , I 2V I — 2V 

I ranspose, x + y — 5 = — = z 

3 3 3 

Since x and y must be integers and 5 is an integer, 

must be really an integer, although of fractional 

3 
form, for one side of an equation cannot be integral and 
the other fractional, that is, y must have such a value that 

2- will reduce to an integer. 

3 

Say, 
then 1 - 2y = $m or 23; = . — 3/;* 

2 

which is still fractional in form. 

To avoid this repetition of fractional form, we have 
recourse to a simple process, based on the truth that an 



I — 

3 


2V 


-- m (an integer) , 


1 — 


2 V - 


$m or 2y = _ - 




y = 


1 - $m 



82 Algebra. 

integer multiplied by an integer will give an integral 
product. 

If — — — ^ is multiplied by an integer, that will make 

the coefficient of y one greater than some multiple of 3, 
the repetition of fractional form will be avoided. 

Observe that this multiplication does not affect our 
result, for we are seeking any integral value not a par- 
ticular one, for 21 . 

3 

To be as simple as possible, the smallest number that 
will suffice for our purpose is chosen for a multiplier ; in 
this case evidently 2, for 2 X 2v = \y, and 4 leaves a 
remainder 1 when divided by 3. 

Then/ 1 - 2 ^ 2 = 2 ~W - -y+^—JL. Since 
V 3 / 3 3 

y is an integer we need only consider the apparent fraction 

~ ? , which we will equate to some integer, say n ; then 
3 

— '—2- = n, 2 — y = $n, y = 2 — 3 n, which is cleared of 

3 
fractions. 

Substituting this value of y in terms of the integer n 
(whose actual value we do not yet know) in the original 
equation, 

3* + 5 (2 - sn) = 16 
3 x + 10 — 15W = 16 
3# = 6 + 15W 
x = 2 + 5». 

Now we have two condition equations to limit the value 
of n. 

x = 2 + 5 n (1) 

and y = 2 — 3 w (2) 






Indeterminate Equations. 83 

Remembering that x and y must be positive, whole 
quantities, we can set limits for n. 

In (1), plainly n may be any positive number from 1 to 
00, but no negative number, because if w = — i,x = — 3, 
which violates our condition. Clearly any other negative 
number will make x negative. 

In (2), n can be any negative number from — 1 to — 00 , 
but no positive number, for a like reason. 

Therefore n, to satisfy both (1) and (2), can be neither a 
positive nor negative number; it can then be only o. 
Hence, if n = o, x = 2 (from (1) ) and y = 2 (from (2) ), 
showing that x = 2 and y = 2 is the only solution for 
3^ + 5v= 16, if limited to positive integers. 

Take another example. 

Solve in positive integers 8 x+ 5 y = 7 4. 

Divide by 5, x + ;y + — =14+-. 

5 5 

Transpose, x + y- I4 = ±-3^= 4 ~ 3* 

5 5 5 



8 - 6„v 3 - x 
= 1 - x + 3 

5 5 

11 (an integer), 3 - x =5», * =3 - 5;*. 



m 



From (1), n can be - 1, o, + 1, +2, etc., anything 
greater than — 2, that is «> — 2. 

From (2), n can be o or any negative number, but no 
positive number. Hence, from both (1) and (2) n can be 
— 1 or o. 



8 4 



Algebra. 



n = — i, x = o, y = 2. 
n = o, x = 3, y = io« 



That is, there are two solutions in this case. 

Art. 114. It is sometimes desirable to know numbers 
with predetermined remainders when divided by given 
numbers. Suppose, for instance, it is required to find the 
least number which when divided by 3, 5, and 6 leaves 
respectively the remainders 1,3, and 4. 

Let x be the required number. 



Then 



some integer, because by the conditions 



of the problem, if 1 be subtracted from x, 3 will exactly 
divide it. 



Also 
and 

say, 



x ~ 3 
5 

x — 4 
6 

x — 1 



= an integer . 



= an integer 



m (an integer) 



x = $m + 1 



• (3) 



(1) 



Substituting this value of x in (2) 



Sm + 1 - 3 _ 



= an 



integer, that is, ■ = an integer. 

Multiply 3m — 2 by 2 (to make coefficient of m one 
greater than a multiple of 5, as explained in last article). 



x m — 2 6 m — 4 m — 4 • . 

£ x 2 = — = m + ■ ± = an integer. 



Indeterminate Equations. 85 



Since m is an integer by supposition, — must be 

an integer, say — = n (an integer). 

m — 4 = 5 n. 
m = $n + 4. 

Therefore, from (1), 

x = 3 (5« + 4) + 1 = 15 w + J 3 • • (4) 

Substituting this value of x in (3) 

157* + 13 - 4 _ l$Jl + 9 _ 5 m + 3 _ „ „ , T , ** + 1 

- — — — ^ TO -t- 1 -t" 

662 2 

= an integer. 
Since 2 n + 1 is integral, must be an integer ; say 

= 5 (an integer) ; n = 2s— 1. 

2 

From (4), * = 15 (2 ^ - 1) + 13 = 305 - 2 . . (5) 
From (5), the least value 5 can have, that will give x a 
positive value, is 1 (s cannot be o, why?). 
If 5 is 1 in (5), x = 28, the required number. 

Verification, — == 9 + remainder 1 
3 

f— ■ • 

The principle applied is this ; that if any number of 
equations (each one expressing one condition for the 
unknown quantity) be combined, the resulting combination 
equation contains all the conditions expressed by the 
component equations. 



86 Algebra. 

Thus when the value of x in (i), expressing the condi- 
tion that x is divisible by 3 with a remainder 1, is substi- 
tuted in (2), the resulting value of x in (4) contains also 
the condition expressed by (2), that is, the x in (4) is not 
only divisible by 3 with remainder 1, but also divisible 
by 5 with remainder 3. 

When this value of x from (4) is substituted in (3), the 
resulting value of x in (5) contains all three conditions, 
hence, by giving s any integral value we please, a value of 
x will result from (5) that will fulfill the three require- 
ments of the problem. Since we want the least value of 
x that fulfills these requirements, we choose the least 
value of s, which is 1. 

EXERCISE XVII. 
Indeterminate Equations. 

Solve in positive integers : 

1. 2 x + 11 y = 83. 

2. \x + 5 y = 92. 

3. %x + \y = 53. 

4. 2 x +3 y = 25. 

5. 12* + i^y = 175. 

6 f* + 3:y+5z = 44. 
Xsx + sy + jz = 68. 

7. Divide 89 into two parts, one of which is divisible 
by 3 and the other by 8. 

8. What is the smallest number which gives a re- 
mainder 4, when divided by 5 or 7 ? 

9. In how many ways can 300 lbs. be weighed with 
only 7 and 9 lbs. weights ? 

10. Divide T 7 T 9 T into two parts, having respectively 
denominators 13 and 9. 



Indeterminate Equations. 87 

11. A wheel with 17 teeth meshes with awheel having 
13 leeth. After how many revolutions of each wheel will 
each tooth occupy its original position ? 

12. How many times each must a 7-inch rule and 
a 13-inch rule be applied to measure 4 feet, using both 
at the same time ? 



CHAPTER VIII. 

QUADRATIC EQUATIONS. 

Article 115. An equation containing the second and 
no higher power of a quantity, is said to be a quadratic 
equation in that quantity. Write three quadratic equations. 

Quadratics in Single Unknown. 

Art. 116. When the quantity whose square is involved 
is a single variable (like x or y) and no other variable 
enters the equation, it is a simple quadratic of one un- 
known quantity. 

The general form is a z 2 + bz + c = o, where a, b, and 
c are constants. 

Kinds of Simple Quadratics. 

Art. 117. If b is zero in a z 2 + bz + c, the resulting 
equation, az 2 + c = o, is called an incomplete or pure 
quadratic; the equation az 2 + bz + c is called a com- 
plete or affected quadratic. 

Roots of a Quadratic. 

Art. 118. If in the equation az 2 + bz + c — o, we 
substitute for z, the value 



b + Vb 2 - 



\a c 



or 



2 a 



thus 



- b± Vb 



2 a 



a_cY 



+ b 



- b -Vb 2 - 


■ 4ac, 


2a 


'- b ± Vb 2 


— 4<jc\ 



) + c = o, 



we get — c + c = o. Verify, 

88 



Quadratic Equations. 89 

These two values of z, which when substituted for z 
make the two sides of an equation identical (or satisfy 
the equation, as it is said), are called roots of z for this 
equation. Formulate a general definition for the roots 
of an equation. 

Solution of an Incomplete Quadratic. 
Art. 119. The equation ay 2 + c = o can be put in 

the form y 2 = — £_ by transposing c and dividing by a. 
a 

What advantage for solution arises from this operation ? 

Since the solution of the equation is the finding of the 

value or values of y that will satisfy it, that is, its 

roots, how would you complete the solution? Formulate 

rule. 

Solution of Complete Quadratics. 

Art. 120. What would above solution suggest as to 
first steps in the solution of the equation 

ax 2 + bx + c = o? 

By Binomial Theorem (x + m) 2 = x 2 + 2 mx + m 2 = n 2 
say, which is a general form of a quadratic equation in x. 
Observe the relation between the third term m 2 and the 
coefficient of x. Suppose this expression to be put in 
the form of an equation, thus : 



How would you restore the form of a perfect square 
to left hand member without altering the truth of 
the equation ? Complete the solution and formulate a 
rule. I 



9o 



Algebra. 



Geometrical Illustration. 



Art. 121. 



AB = AC - BC = n - m (see Fig. 4.) 
n 2 = CEFA = BGHA + BCDG + GKFH 

+ DEKG = x 2 -\-mx + mx + m 2 
= x 2 + 2 mx + m 2 . 



c 


D 


E 




/ 












B 


mx 


m 


m 2 




I 


X 


G 


m 




\ 




X 2 


X 


mX 






V 











Fig. 4. 



LITERAL AND NUMERICAL EQUATIONS. 

Art. 122. Equations such as we have considered, 
involving letters as coefficients, are called literal equa- 
tions ; if the coefficients are numbers, they are said to be 
numerical equations. Write three numerical quadratic 
equations. 

Solution. 

Art. 123. What are the essential differences between 
the use of letters and the use of numbers in solution ? 



Quadratic Equations. 91 

Observe carefully the following steps; by Binomial 
Theorem, 

(pc - 5) 2 = x 2 - 105; + 25 = 49, say, . (a) 
transposing, 

x 2 — io# = 24 or x 2 — 101 — 24 = o, . (b) 

compare ax 2 + bx+c = o. . (c) 

An examination of (a) shows that a simple extraction 
of the square root as in case of pure quadratic, will give 
the values of x, thus: 

x — 5 = ±7, whence x — 12 or — 2. 

Plainly then, to solve a quadratic like (b) or (r), we 
must put it in (a) form. 

What must be added to x 2 - 10 x in (b) to restore 
form (x — 5) 2 ? What relation does this added quantity 
bear to coefficient of xl 

Again, (x + |) 2 = x 2 + * .r + * = 1, 

x 2 + %x = f, transposing. 

Complete solution. 

Solution by Factoring. 

Art. 124. Solve .v 2 - 7.V+ 10 = o 

x 2 - 7 x = — 10 [transposing] 

x 2 - 73: + (£) 2 = (|) 2 - 10 = I [completing square] 
x -i - ± | 

whence x = £ - + | = 5 

or ^ = t._3 = 2 

But .v 2 — j x + 10 may be resolved into the factors (x - 2) 
and (x - 5) ; hence (x - 2) (pc - 5) = o. 



9 2 



Algebra. 



'£> 



By inspection, it is plain that if either 2 or 5 is substi- 
tuted for x, the equation is satisfied; for (2 — 2) (2— 5) 
= (o)x( — 3) = o (since any finite quantity multiplied 
by o equals o), and (5 - 2) (5 - 5) = (3) x (o) = o. 

Hence, the roots of the equation are 2 and 5, as found 
above. This is known as solution by factoring and is of 
advantage when the equation is easily factorable. Factor- 
ing may be accomplished in any case as follows : take 
same equation, x 2 — J x + 10 = 0, to complete the square, 
\ 9 must be added to x 2 — 7 x, write then the equation thus, 
(a) x 2 — 7 x + \ 9 — I = o, which does not alter the value, 
merely the form of the equation. 

The last equation can also be written (x - |-) 2 - (|) 2 = o, 
which is the difference of two squares, hence factorable 
into the product of the sum and difference of the square 
roots of these terms, i.e. (# — £ + $) (x — \ — |) = o, or 
(x - 2) (x - 5) = o, as before. 

To get a general result, let us take ax 2 + bx + c = o. 

Whence x 2 -\ — x + — = o. 



To make complete square of x 2 -1 — x, we must add 



<:/ 



— , hence x 2 + - x + 
4a 2 a 



or Ix + — r — ( -— - ) 2 = o, a difference of 

\ 2a J \ 20, J 

two squares hence equals product of two factors. 



\ 2a 2a J \ 2a 2a ) 

. „_ - b + V b 2 - \ac ^ - b - ^ b 2 - 4ac 

, . x — or — 



2a 2a 



Quadratic Equations. 93 

Solution by Substitution. 

Art. 125. Since in the above solution letters alone 
were involved, they may evidently stand for any numbers 
we please. 

Let us compare the two equations just used, 

x 2 — 7^+10 = and ax 2 + bx + c = o ; 

we may if we choose, say that a = 1, b = — 7, and c = 10 ; 
ax 2 + bx + c = o then becomes x 2 — 7 x + 10 = 0. 

The two values of x found above, if a = i,b = — 7, and 
c = 10, will then plainly become the values of x for x 2 
- jx + 10 = o ; i.e.. 



— ^£. becomes x = *-^± ^2 — 

2a 2 

7 ± 3 = 5, or 2 as before. 



Why is ax 2 + J* + c = o called the general form of 
the quadratic equation ? 

EQUATIONS INVOLVING RADICALS AND REDUCIBLE 
TO QUADRATICS. 

Art. 126. The process of removing the radical ex- 
pressions from an equation by squaring, not infrequently 
introduces extraneous roots ; it is therefore necessary to 
verify the results carefully in each case, thus : 



Vjc + 3 + V43; + 1 = \/io* + 4 . . . ( a ) 
whence squaring, 
x + 3 + 2 ^(x + 3) (4.x + 1) + 4X + 1 = iox + 4. 
2 v (x + 3) (4X + 1) = 4-5X [collecting] 

i6x 2 + 52 # + 12 = 25X 2 [squaring again] 



2 S 2 ^ 
x 2 - 2 



~~ = $ (&)> [collecting and dividing by 9] 
whence x = 6 or - f . Complete the solution of (b). 



94 Algebra. 

Substitute 6 in (a), we get 3 + 5 = 8, hence 6 is a root 
of (a). 

But if we substitute — —in (a) we get ^ +" — = ! 
9 3 3 3' 

which is false. It is plain, however, that if the second 

2 

9 



radical is negative, the value - — would satisfy, for 



3 3 3 



Have we any right to use negative sign with V \ x + 1 ? 
Why ? — § is then a root of 

V x + 3 _ V 4 # + j. = v 7 ^ # + 4 . . ( c ) ; 

squaring (c) 

x + 3 - 2"^ (# + 3) (43: + 1) -{- 4 x + 1 = 10 x + 4. 



Whence, ~ 2\/(tf + 3) (4.x + 1) = 53c. 

Squaring, i6x 2 + 52 x+ 12 = 25 f as before. Why? 
Hence both (a) and (c) lead to same quadratic whose 
roots are 6 and — f . 

Emphasis then must be laid on the examination of all 
roots, where we are required to square terms of the equa- 
tion. 

EQUATIONS OF HIGHER DEGREE SOLUBLE AS 
QUADRATICS. 

Art. 127. Recalling the definition of a quadratic equa- 
tion, it will be observed that the term quantity means not 
necessarily a simple letter, nor even a monomial. 

We may extend the definition thus : Any equation involv- 
ing only such expressions in the variable as may be collected 
into two exactly similar groups, whose exponents shall have 
the ratio of 2 to 1, maybe solved or at least partially solved 
as a quadratic. 



Quadratic Equations. 95 

Solution. 
Art. 128. 

ofr + x^ — 20 =• o. 

Let x\ = y, and hence expressed in quadratic form 

y + y 2 — 20=0 Complete solution. 

Again, 5c 2 — 2x + 6 ^ x 2 - 2 x +5 = 11 

may be written, (x 2 — 2x + 5) 1 + 6 (x 2 — 2# + 5)* = 16. 
Since the exponents of the similar groups (x- - 2.x + 5) 1 
and (x 2 — 2# + 5)* have ratio of 2 to 1, 

let (x 2 - 2x + 5)% = y. 

Then the equation becomes, 

y 2 + 6y = 16, 
whence, 3/ = 2 or - 8 ; i.e., (x 2 - 2x + 5)* = 2 or -8, 
whence, x 2 - 2 x + 5 = 4 or 64 ; 

x 2 — 2 x = — 1 or 59, 
* - x, 1, (x + 2^ / ~i5) or (1-2 ^71). 
Again, 4X 4 — 12X 3 + $x 2 + 6x — 15 = 0, 

arranged, 4.x 4 — 12 x 3 + 9X 2 - 4X 2 + 6x = 15, 

or (2 x 2 - 3 x) 2 - 2 (2 x 2 - 3a;) = 15, 

plainly of quadratic form. Complete solution. 

The grouping and arrangement of such equations is a 
pure matter of judgment and ingenuity and can be sub- 
jected to no general rules. 

Solution of Higher Equations by Factoring. 

Art. '129. Any equation of the form x n + bx n ~ l + 
ex*- 2 + ... =0 whose left hand member can be 
resolved into factors of degree not higher than the second 
may be solved completely. 



g6 Algebra. 

Art. 130. Solution. 

x 3 — 6 x 2 + 1 1 x — 6 = 0, may be written 

x 3 — 6 x 2 + 12^-8-^ + 2 = 0, 
or x 3 — 6 x 2 + 12 x — 8 — (x — 2) = o, 

or (x - 2) 3 — (x — 2) = o, 

which may be factored thus : 

(x - 2) [ (x - 2) 2 - 1] = o, or (x - 2) [ (x - 2) - 1] 
[ (x — 2) + 1] = o, or (x — 2) (x — 3) (x — 1) = o. 

Plainly, if x = 2 or 3 or 1, the equation is satisfied. 
These values arise from setting the factors successively 
equal to o. 

Again, (x — 1) (x — 2) (x 2 — \x — 5) = o. 

Clearly any value of x that will reduce any one factor 
to zero will satisfy the equation, provided it does not 
make another factor infinite. These values will result 
from the solution of the three equations 

x — 1 = o 
x — 2 = o 
x 2 - 4 x - 5 = 0. (Why ?) 

x — 1 = o gives x=i, x — 2 = gives x = 2, and 
x 2 — 4X— 5 = gives x = + 5 and x = — 1, hence 
1,2, — 1, + 5 substituted for x in the original equation 
identically satisfy it. 

CHARACTER OF ROOTS. 

Art. 131. Since the solution of an equation with 
literal coefficients gives general results, we may derive 
useful information from a study of the roots of the equation 



, , , • . - b + v b 2 - 4 ac 
ax 2 + bx + c = o, i.e., x = 



2 a 



_ 1 _ Vp _ 4 ac 

and x = — 



2 a 



Quadratic Equations. 97 

If we examine these roots carefully, we see that the 
difference in their values arises from the addition of 



vV - \ac 

2 a 



in one case and its subtraction in the other from 



the same quantity . Put this statement into form of a 

2a 

general rule. 



If, then, v#2 _ ^ ac j s Qj trie r0 ots will both be the 
same, hence this is the condition for equal roots, which 
has important applications in other branches of mathe- 
matics. Put this condition into a rule. 

Again, ^b 2 — \ac determines whether the roots shall 
be rational, real, or imaginary. 

What condition must b 2 — \ac fulfill in each case? 

Illustration. 

5 tt 2 + 6 # = 8 can be written $x 2 + 6x — 8 = 0. 

Compare, 5 x 2 + 6 x — 8 = 

ax 2 + bx+c=o. 

If a = 5, b = 6, c = — 8, the results of the solution for 
ax 2 + bx + c = o, will be those for 5 x 2 + 6 x — 8 = o. 
Why? 

Then, 

- b + V> - \ac - b - ^b 2 - ±ac 
x = - or - — 



20 2 a 

becomes, 



6 + ^36 + 160 - 6 - v^6 + 160 . 4 
^ — or ^ j -(?m ft or -2. 

10 10 c 



Now b 2 — 4-ac = 196 = 14 2 is a perfect square, hence 
4 
5 



4 , 

the rational roots — ancl — 2 « 



Algebra. 



Illustration. 



In 2>y 2 — %y + 10 = o, a = 3, b = — 8, c=io. Here 
b 2 - \ac = (- 8) 2 - (4 x 3 X 10) = 64 - 120 = - 56. 

Hence roots are 



+ > V ~*4 and 8 - 2N/37 



6 




6 




Which condition 


prevails here ? 






Illustration. 








6 + -51 = 6t\ 

2 


Here a = 6, 5 = 


2 


-6. 


Hence 








i2 -^ = (-f) 2 


- (4 x 6 x - 6) = 


■ «■ + 144 

4 


601 
4 


Write roots and state condition. 








EXERCISE XVIII. 








Quadratics. 






Verify results in 


every case. 






1. 2^-27 = 9 


x-x 2 + $. 3. 5 


^-3 = 10/- 


-3^ 


2. :y 2_ 5;y _ 24 


= 0. 4. s = 


= J g/ 2 (solve 


for t). 


5. X. 
21 


2 
= — (Pendulum formula). 




6. 5 = 7// + J a/ 2 . 


»• 5- 


r 2 = 8 x. 





8. Solve x 2 + 6 x = o by general rule, and then show 
how this equation can be solved by shorter method. 

9. Prove from a solution of the general equation 
ax 2 + bx + c = o, that if c = o one root is o, and hence 
derive a rule relative to the absolute term. 

10. 2 y 2 -s v = 3 y+234. 



Quadratic Equations. 99 

11. 3 y 2 -8 y=2 y (3; -4) + 9. 

12. What must be added to P 2 - 5 P to make the 
expression a perfect square ? 

13. Make 9 a 2 x 4 + 12 a# 2 a perfect square. 

14. 3f+5»-4 = i 2 -2x + 3. 

15. 6 + 5 / = 6 t 2 . 

16. *2 +2 «= (*-i) 3 -*+24 a 

^+ 2 

17. i2^-a-2o c 2 = o. 

18. 2 ^ + 5 _ 2 *+7 _3 = Q> 
3^-2 3 x-4 4 

19. 8 #-x 2 -i2 = o. 

20. ( 2 y- 3 ) 2 = 6(y+i)- 5 . 

21. I* _ 3_*?+2 = o. 
2 4 

22. 15 y 2 — 7 j- 2 = 0. 

23. x 2 + (rn + n) x + mn=o. 

oa 3 , m , am or 2 

24. -^ + 2 = — / — — . 25. 3m 2 -4m-io = o. 

a n 2 w 



EXERCISE XIX. 

Solve following examples by the process that seems 
most expeditious to you. 

1. 6 x 2 — 7 x+ 3 ' = °- 2 - # 2 — 21 x+ 104 = 0. 

3 - (x-5) 2 + x 2 = 16 (.T + 3). 

Q 

4. 5 # 2 # + 4 = 0. 5. x 2 — 13 x=— 42. 

3 

6. £±J +(x+i)(x+2) = o. 

17 

7. # + =2 (#—2). 

# — 6 

L.OFC* 



100 Algebra. 



8. 


x— (a + b) x= — ab. 


9. 


3 1 _7 
4 x 2 6 x 2 3 


10. 


x 2 + bx + a =bx (i — bx) 


. 




11. 


x (io + x) = — 21. 


12. 


* 2 = *. 


13 


-I - I = 6. 

x 2 x 


14. 


x 2 + 9 # = 8.5. 


15. 


4.05 x 2 — 7.2 x = 1476. 






16. 


6 + 5 x 3 x-4 , 
4 (5 - *) 5 (5 + x) 


5- 
2 5 


7 x 89 _ Q 
— x 2 105 


17. 


25 x = 6 x + 21. 


18. 


x+22 — 6 x 2 = o. 


19. 


2 x 2 — ax + 2 ox = a&. 


20. 


2 x 2 — 7 x + 3 = o. 


21. 


2 , 2 x 2 , 
3X 2 + 3= +24# 

3 


22. 


^ , 2 _J3 

x + 3 x + 6 20 


23. 


-° +i = 3*. 

x+ 2 


24. 


1 2 

2 +x= — • 



25. b(a-x) 2 =(b-i) x 2 . 
Verify results in each case. 

26. The number of square inches in the area of a 
square exceeds the number of inches in its perimeter by 
32. What is the area? 

27. A hall can be paved with 200 square tiles of a 
certain size ; if each tile were one inch longer each way 
it would take 128 tiles. Find size of tile. 

28. A wheel driving a drum makes 15 less revolutions 
than the drum in rolling up 440 feet of rope. If they 
were each 2 feet more in circumference, the wheel would 
make 10 less revolutions. Find circumference of each. 

29. A lever, cut from a bar weighing 4.2 pounds per 
foot, balances at a point 2.3 feet from one end if 54 pounds 
is suspended from that end. Find length of bar. 

30. When a lever AB is supported at its C.G. it is found 
that a weight W at A will balance 2.5 pounds at B; but 
W at B requires 19 pounds at A to balance it. Find W. 



Quadratic Equations. 101 

EXERCISE XX. 

1. o<? - x* = 56. 

2. x* + x% = 756. 

3. x + 16 - 7^# + 16 = 10 — 4^x+ 16. 

4. ^x + 12 + Vx + 12 = 6. 

5. jc 6 + 7^ 3 = 8. 

6. 3^- 3*-* = 8. 

7. jc 4 + 2 x 3 - 3 x 2 - 5x + 4 = o. 

8. 2 x 2 + 3 jc + 1 = — 



2 x 2 + 3 x 



9. 3 x (3 - *) = 1 1 - 4 V * 2 - 3 * + 5- 

10 . (* + i)>_ 4 (* + i)- s . 



*" - 3 . 3* _ 13 



11. 1 0- + 



# # 2 - 3 2 



12. -y 4 + 2y 3 + 5^ + 4/y = 60. 



13. z 2 — 52; + 2^ z 2 - 52 - 2 = 10. 

14. w 4 + 211 3 — n 2 — 4.11 - 96 = o. 

Remark. It is evident that the number of equations that can 
be solved thus is very limited ; the general solution of third and 
fourth degree equations cannot be considered here. 

Determine all roots of the following equations : 
1. x + v x = 4^ — 4 Vx. 

2. V \y + jj + Vy + ! _ 4 = o. 



3. Vjc 4 1 |(, V +l)-i=2. 



4. ^4^4 ! _ \/ x 4 3 = v / #_ 2 . 



102 Algebra. 



6. V x _ 2 + V3 + x~ ^19 + x 



o. 



6. (1 + 2x)% - ^4 + x + ^3 - x = o. 



7. (43: —2)^+2 V2 — a; — ^14 — 4x = o. 

8. ^{x - 1) (x - 2) + v^ _ 3) ( x _ 4 ) = 2. 



9. v# + 3 - ^x + 8 = 5^. 
10. ^ + ^ - 20 = o (discuss roots thoroughly). 



11. ^ / z + 5 + ^ / 3z + 4 = x/ i2z+ 1. 

12. v^ + 9 + ^49 - x = ^x+ 16. 



MAXIMA AND MINIMA. 

Art. 132. It is often desirable to know how large or 
how small an expression may be made by altering the 
unknown quantity involved in it within rational limits. 

It might for instance be required to find the largest 
rectangular beam that could be cut from a cylindrical 
log of known diameter, or how to divide up a line so 
that its parts would inclose the largest area when made 
sides of a figure of a certain general form. Say, for 
example, it were required to find the largest value the 
expression 5 +24X — gx 2 can have, if x varies within 
real limits. 

Say the value of this expression is m, where m has 
a changing value, of course, as x changes. 

Then, 5 + 24a;- gx 2 = m. 

Now since the value of x depends upon the value of m, 
because we want to find what x is, when m is the largest 
possible, it is plainly desirable to solve the above equa- 
tion for x, so we can see in its simplest form just the kind 
of dependence that x has upon m. 



Quadratic Equations. 103 

Transposing, then changing the sign, and completing 
the square : 



9 


x 2 - 


• 24X 


= 


5 - m 








X 2 


~ f* 


= 


5 - m 
9 








" "3 X 


9 

X — -3 


= 


21 — m 
9 








± V21 


- 


m 






X ■■ 




/I J- \A 


3 






1 - 


- 7» 



whence, 



By an inspection of this value of x, it can be readily 
seen that if m has a value greater than 2 1 , the expression 
under the radical (21 — m) will be a negative quantity. 

For example, 



if m = 22, then 21 — ;w = — 1 and x = 4 ± V — 1 

3 

an imaginary value. As only real values of x can be con- 
sidered, clearly any value of m that is greater than 21 is 
impossible. 2 1 then is the largest value m can have and 
is called the maximum value of m. If m is 21 (its largest 
value), then the radical 21 — m — o and x = f. That is, 
the value f for x makes the expression 5 + 24.x — gx 2 as 
large as it can be. m can evidently be anything less than 
21, and hence the expression 5 + 24V — 9V 2 has no 
minimum value. Again, let it be required to divide any 
number, say a, into two such parts, that their product shall 
be a maximum. 

Let x = one part. 

Then, a — x = the other 

and ax — x 2 is to be a maximum. 



104 Algebra. 



Let ax — x 2 = m 

2 , a 2 a 2 — 4 m 
x 2 — ax -\ — =- — 

4 4 

# = ± Va 2 — \m 

2 ) 

2 

x = a ± \/ a 2 — 4.m 



4 m cannot be greater than a 2 , or x will be imaginary, 

a 2 
that is, m cannot be greater than — , and for this maxi- 
mum value of m, x = ~~, hence the product of the parts 
is greatest when they are equal. 

EXERCISE XXI. 

Find maximum or minimum values of following 



ressions : 








1. x 2 — 6 x + 


i3- 


5. 


X J — x — I 


X 2 — X + I 


2. 3 + i2j - 


9^ 2 . 


6. 


I I 


2 + X 2 — X 


3. X ~ 6 . 




7. 


x 2 + 3 x + 5 


X' 






#-' + i 


4. 4 x 

(X + 2) 2 




8. 


12 + X 2 — 2<2#. 



9. x 2 — io# + 35. 

10. Find greatest rectangle that can be inscribed in 
circle of radius 10 inches. 

11. Divide a line 12 inches long into two parts such 
that their product shall be a maximum. 

12. Find length of the sides of the largest rectangle 
having perimeter 16. 



Quadratic Equations. T05 

13. If you were three miles from shore in a boat, and 
could row four miles per hour and walk five miles per hour, 
and wanted to reach a point on the beach five miles away 
in the shortest time, where would you land ? 

14. Find maximum value of x for real values of r in 
equation 1 = xrd + x 2 — r 2 . 

15. 2 x 2 — $yx — Sy 2 + 18 = o, find minimum value of 
y, if y is always positive. 

EQUATIONS CONTAINING TWO OR MORE UNKNOWN 

QUANTITIES OF A DEGREE HIGHER THAN 

THE FIRST. 

Art. 133. Equations involving more than one un- 
known quantity and of certain forms can be readily solved 
by special methods, usually reducible to the quadratic 
solution. 

Such equations may be classified ; first as homo- 
geneous equations of the second degree involving two 
unknowns, i.e., two simultaneous equations such that the 
terms containing the unknowns are all of the second 
degree in both equations. 

Thus, x 2 + 2xy - f = 28 . . . . (a) 

3 x 2 + 2 xy + 2 y 2 = 7 2 . . . . (b) 

All such equations may be solved by substituting y = mx 
or x = ny. 

Solution. Then (a) becomes x 2 +2mx 2 — m 2 x 2 =28 
(sub. y = mx) and (b) becomes 3 x 2 + 2 mx* + 2 m 2 x 2 = 72. 

Whence, from (a) 

and from (b) 

28 72 ^ 22 m x 

• m — = 1 or ;;r = «- 

1 + 2 m — m 1 3 + 2 m + 2 m 2 32 32 



x- 




I 


+ 


2 m 


- 


m 2 


X 2 


_ 






72 




















3 


+ 


2 m 


+ 


2 »r 






7: 


2 




• or m 2 - 



106 Algebra. 

Whence, m = — or -$- 

2 16 

and .*. x 2 = - = 1 6, 

i + i-i 

# = ± 4 y = ± 2, etc. 

In special cases briefer methods may be employed 
which depend entirely upon the ingenuity of the solver. 
As in same equations ; 

add the equations; 4 x 2 + 4 xy + y 2 = 100, 
whence, extracting the square root, 2 x + y = ±10, 
whence, y= 10 — 2ior- 10 — 21 

Substitute first value in (a), and a quadratic in x results. 
Finish solution. 

Art. 134. When one equation is linear (of first degree), 
the method of substitution is generally most effective, as 
indicated at the conclusion of last article. 

Solution. 

{25 + 3/= 10, whence 5 = — 
* (s + t) = 25 
.*. / ( — + / j = 25, (substituting in 2d equation) 

- t 2 + 10/ = 50 
t 2 - lot = -50. 



Whence, t = 5 (1 ± ^-1) 

and 5 = 5 (~ T ^ 3 v -i\ verify. 

2 

The exercise of a little judgment and ingenuity will 
often simplify the solution of problems of this kind also. 
It may be said as a general remark, that there is a large 



Quadratic Equations. 107 

field in algebra for the application of legitimate artifice to 
shorten labor of calculation. For instance, 

x ~ 3 y + 9 = ° M 

xy — y 2 + 4 = o (b) 

Transposing and squaring in (a) x 2 — 6 xy + 9 y 2 = 81 
Multiply (5) by 8 and add 8 xy — 8 ^ 2 = — 3 2 

fl2xy+ ;y 2 = 49 
Extract square root x + y = ± 7 . . (<:) 

Subtract (a) from (c) x - 3 y = — 9 

\y = 16 or 2, etc. 

Lose no opportunity to apply such methods, but 
remember they are worth while only when quickly 
observed. 

Art. 135. When the equations are both symmetrical, 
they may often be readily solved by substituting 

x = it + v. 
y = u — v. 

Equations are symmetrical when the unknown quantities 
may be interchanged without affecting the equation, as, 

x 4 + y 4 = 706 (a) and x 2 + 3 xy + y 2 = 125. 

x + y = 2 (b) and x 5 + x 4 y + xy 4 + y 5 = 1020. 

Making above substitution in (a) and (b), 
(u + v) 4 + (u - v) 4 = 706 
or 2 u 4 + 12 u 2 v 2 + 2 v 4 = 706. 

Whence, n 4 + 6u 2 v 2 + v 4 = 353 . (a) 

and ^ + ^ + 2^-^ = 2. Substituting in (b) 

Whence, u = 1. 

Substituting m = 1 in (a), 1 + 6zr-(- v 4 = 353. 



io8 Algebra. 



Whence, 


v = ± 4 or ± V — 22. 




Whence, 


x = u + v= i±4 or, 
i± ^ _ 22 ; y = u — v, etc. 




Again, 


X 5 — ^ 5 = 211 


. (a) 




# — y= i . . . . 


• (b) 



Divide (a) by (ft) ; x 4 + x 3 v + x 2 y 2 + xy 3 + y 4 = 211 
raise (5) to 4th power, x 4 — 4 x 3 v + 6 ^y — 4 xy 3 + y 4 = 1 



subtract, 5 x 3 y - 5 x 2 y 2 + 5 xy 3 


= 210 


x 3 y — #y + xy 3 


= 42 


square (b) and multiply by xy, 




x 3 y — 2 x 2 y 2 + xy 3 


= xy 


subtract, x 2 y 2 + xy 


= 42 


complete square, 




x 2 f + xy + J = if *.. 




T 2 • 

^ + i = ± > xy = 6 or — j . . 


. (c) 






Multiply (c) by 4 and add to square of (b) ; 

x 2 — 2 xy + y 2 = 1 

4xy = 24 or — 28 



ar + 2 xv + y 2 = 2 5 or — 27 

x + 3/ = ±5 or ± 3 V_ 3 
x — y = 1 (b) 

which indicates how. a general solution may be varied in 
special cases. 






Quadratic Equations. 109 

Art. 136. One equation may be divisible by the other, 
as, 

Solution. 

x 4 + x 2 y 2 + y 4 = 931 (a) 

x 2 + xy + y 2 = 49 (b) 

x 2 - xy + y 2 = 19 (c) 

2X y = 30; xy = 15 . . . . . . (d). 

Divide (a) by (ft). 

Subtract (c) from (b). 

Add (d) to (b) and subtract (d) from (c). 

x 2 + 2 acy + y 2 = 64 whence, x + y = ± 8 

x 2 — 2 xy + y 2 = 4. x — y = ± 2, etc. 

Again, x 3 - j 3 = 7 jcj (a) 

x — ^ = 2 (5) 

Divide (a) by (b) ; x 2 + xy + y 2 = % xy. 

or # 2 — $ — ^ + y 2 = o (c) 

2 



Divide (r) by (> 2 ). 

— - -^(I_)+ 1=0 quadratic in (— ) 

f 2 \yJ L \y/i 

gH©-- 

Whence, - -i = ± 1, - = 2 or ',. 

y 4 4 v 

Whence, x = 2 y or \ y. Complete. 



no 



, Algebra. 



EXERCISE XXII. 

Compose examples of each type indicated and solve 
them. 



3. 



5. 



x 2 ,'f_ I 9 
y x 6 

x y 6 


2. . 


2 
X + y =— • 

6 


x'y 2 — 1 6 xy + 6o = o 

x + y =7. 


4. 


; x 3 + y 3 = 1 - 3 3: 

'x 2 + y 2 = xy + 3 7. 


x 2 + y 2 = axy 
x+y =bxy. 


6. 


x 2 -3xy + y 2 =$ 
x* + y* =2. 



7. 



x 2 + y 2 + 5 ^x 2 + y 2 = 50 
x 2 -y 2 =7. 



8. 



3 x- 2 — y~ 2 = 1 
Sx~ 2 -(xy)- 1 + 2 y~ 2 = $. 



x + y x — y _ ic 

x— v x + y ; 
x 2 + y 2 =45 



10. 



11. 



12. 



x 2 + 3 xy + 3 (x — y) 
x 2 + 2 xy — 3 y 2 

x 2 + xy + y 2 = 63 
^ + ^ = ~3- 

£x 2 + £y 2 -6o = o 
£x + iy-5 =0. 



13. 



# 2 + 3#;y = 54 
xy+4y 2 =ii5. 



14. 



1 x* — y— 127 

'x 2 y-xy 2 = 42. 



15. 



x 2 + xy + y 2 = 84 



V 



xy + y =6. 



Quadratic Equations. 1 1 1 



16. 

17. 

18. 
19. 



I x + y = io. 



x* + 
3xz + 

x + 


y 2 -z 2 
Zyz-2 
y — z 


xy = 


21 

18 
5- 


x 2 y*- 
xy- 


- 6 xy 2 = 

-y = 


-9 

2. 





(x 2 + xy+ 2 y 2 =74 
(2 x 2 + 2xy + y 2 = 73. 



20. [*+*y-*s 21. j^-4/-9 

xy — y = 2. (.v^+2_y- = 4. 

22. f *(x + y) = 3xy 

\x + y + x 2 + y 2 = 26. 

23. (*?(*+?) -30 

^ + ^-65 25 ( .ix 2 +.5y-2 =0 

3,^=28. ' ( .ix — .25^ — 3 = o. 



EXERCISE XXIII. 
Problems. 

1. The product of the number 2^3 and 4 # 6 in the 

decimal system is 1 15 368. What is the digit, x ? 

2. The sum of a number and its square root is 42. 
Find the number. 

3. The area of a rectangle is 120 square feet and its 
diagonal is 17 feet. Find length and breadth. 

4. A square and a rectangle have together the area 
220 square yards. The breadth of the rectangle is 9 yards 
and its length equals the side of the square. Find area of 
square. 



112 Algebra 



5. From the vertex of a right angle two bodies move 
on the sides of the angle, one at rate of 1.5 feet and other 
2 feet per second. After how long are they 50 feet apart? 

6. If the sides of an equilateral triangle are shortened 
8, 7, and 6 inches respectively, a right angled triangle is 
formed. Find the side of the equilateral. 

7. About the point of intersection of the diagonals of 
a square as a center, a circle is described ; the circumfer- 
ence passes through the mid-points of the semi-diagonals ; 
the area between the circumference and the sides of the 
square is 971.68 square inches. Find the length of side 
of square (v = 3^). 

8. The fore wheel of a carriage turns in a mile 132 
times more than the hind wheel. If the circumference of 
each were increased 2 feet, the fore wheel would turn only 
88 times more. Find the circumferences. 

9. A cistern can be filled by 2 pipes ; one can fill it in 
2 hours less than the other ; it can be filled by both pipes 
running at once in 1$ hours. Find time for each. 

10. A and B are laying a cement walk. At A's rate of 
work he could finish the job himself in 18 hours ; B lays 9 
running yards per hour. A finishes his portion in as 
many hours as B lays yards per hour. Find amount laid 
by each. 

11. Two cubical tanks have together 407 cubic feet 
contents. The sum of their edges (outside measure) 
= 11 feet 1 inch. Tanks are made of \ inch steel. Find 
amount of steel necessary for them. 

12. A body starts from rest under acceleration of 18 
feet per second, find the time required to pass over the 
first foot ; the second ; the third. 

13. In going 173.25 yards the front wheel of a wagon 
makes 165 revolutions more than the rear wheel, but if 
the circumference of each wheel were 27 inches more, the 
front wheel would, in going same distance, make only 112 



Quadratic Equations. 1 1 3 

revolutions more than rear one. Find circumference of 
each. 

14. Two points, A and B, start at same time from a fixed 
point and move about circumference of a circle in oppo- 
site directions, each at a uniform rate, and meet after 6 
seconds. The point A passes over the entire circumfer- 
ence in 9 seconds less time than B. Find the time taken 
A and B to travel entire circumference. 

15. A reservoir has a supply pipe, A, and a discharge 
pipe, B. A can fill the reservoir in 8 minutes less time 
than B can empty it. If both pipes are open, the reservoir 
is filled in 6 minutes. Required number of minutes it will 
take to fill, if A is open and B closed. 

16. A body is projected vertically upward with a velo- 
city of 80 feet per second. When will it reach a height of 
64 feet ? 

17. A lawn 25 feet wide and 40 feet long has a brick 
walk of uniform width around it. The area of the walk is 
750 square feet. Find the width. 

18. The perimeter of a rectangular field is 184 rods 
and the field contains 12 acres. What are its dimensions ? 



CHAPTER IX. 
LOGARITHMS. 

Article 137. The logarithm of a number is the power 
to which a given number, called the base, must be raised 
that this power may equal the number. 

For instance, take 2 as a base, 

then, 2 1 = 2, exponent 1. 

2 2 = 4, • exponent 2. 

2 3 = 8, exponent 3. 

2 4 = 16, etc. exponent 4, etc. 

Hence, 1 is the logarithm of 2 to base 2. 

2 is the logarithm of 4 to base 2. 

3 is the logarithm of 8 to base 2. 

4 is the logarithm of 16 to base 2. 

Plainly any number, except o or 1, may be selected as 
a base. Why not o or 1 ? 

Art. 138. It has become customary to use 10 as the 
base for logarithms, principally for the reason that 10 is 
also the base of our number systems, both integral and 
decimal, and hence is best adapted for the base of 
logarithms of these numbers. 

Take then a series of powers of 19 ; thus : 

io 1 = 10. 
io 2 = 100. 

IO 3 = IOOO. 

io 4 = 10,000, etc. 
114 



Logarithms. 1 1 5 

Then with the base, 10, the 

log 10 = I. 

log ioo = 2. 
log iooo = 3. 
log 10,000 = 4. 

By inference the logarithm of any number between 1 o 
and 100 is between 1 and 2, of numbers between 100 and 
1000 is between 2 and 3, etc. 

Hence, to represent all numbers, it is necessary to 
employ fractional powers, for instance, the logarithm of 





29 = 1.4624 + 


of 


327 = 2.5145 + etc. 


That is, 


20 = lO I,4 ^ 24+ 


and 


327 = io 2 -s'45+ etc. 



These fractional powers of 10, which we call the 
logarithms of the numbers to which they correspond, are 
found by computation from a series, which is of no 
especial interest here. 

The first logarithms for general use were based upon an 
incommensurable decimal. 2.7 1828 18 + (usually repre- 
sented by e), and known as Naperian logarithms from 
their discoverer, Baron Napier. 

Art. 139. Since logarithms are exponents (usually 
exponents of 10), they obey the laws of exponents; 
namely, in multiplying, exponents, and hence logarithms, 
are added ; in dividing, exponents, and hence logarithms, 
are subtracted, etc. Hence, the following rules : 

The logarithm of a product equals the sum of 
the logarithms of the factors. the logarithm of a 
quotient equals the logarithm of the dividend, 
minus that of the divisor. state the rules for 
powers and roots, from analogy to exponents. 



n6 ' Algebra. 

Art. 140. The use of the base 10 makes it possible 
not only to simplify calculation by logarithms, but also to 
express them in a much more compact tabulated form. 
For example, take the series of powers of 10 again. 



log = 1. + -! \ two digits 

(W= 100 h 

log = 2. + \ V three digits 

[ fio 3 = iooo^l J 

log = 3. + -j \ four digits, etc. 



It is apparent that any number between 10 and 100 
has a logarithm 1 + a fraction ; any number between 
100 and 1000 has a logarithm 2 + a fraction, etc. 

But every number between 10 and 100 is composed 
of two digits in its integral part, for instance 23, 29.375, 
57-5> etc. 

Every number between 100 and 1000 is composed of 
three digits, as 237, 676, 253, 987, 234.2, etc. 

Hence, the whole part of the logarithm of a number is 
always one less in absolute value than the number of 
digits in the integral part of the number. 

This fact may be shown in tabulated form ; 

Between f log * 3 = *-3 6l 7 5 whole part is 1. 
j ilog 67.6 = 1.8209 : whole part is 1. 

10 and 100 & ' - v i t. • 

Uog 98.2 = 1.992 1 ; whole part is 1. 

Between f lo & 2 ^ = 2 ' 3711 ; whole part is 2 ' 
100 and 1000 \ lo Z 595-35 = 2.7769; whole part is 2. 

Uog 802 = 2.9042 ; whole part is 2, etc. 

Art. 141. The whole part of a logarithm is called its 
characteristic, and the decimal part is called its mantissa. 



Logarithms. 117 

Art. 142. Since numbers which have the same figures 
(digits) arranged in the same order, differ from one another 
only by some multiple of ten, and since the logarithms of 
multiples of ten are always' whole numbers, it follows that 
the decimal part of the logarithm remains the same so 
long as the digits are unchanged, no matter where the 
decimal point be placed. 

Example. 23456, 2345.6, 234.56, 23.456, 2.3456, 
.23456, .023456, etc., all have the same mantissa in their 
logarithms. This makes it possible to find the logarithms 
of all numbers from any table of logarithms, as shown 
later. 



Logarithms of Decimals. 

Art. 143. A pure decimal always indicates a fraction 
with a denominator which is a higher power of ten than 
the numerator, hence, since a fraction means a division, 
by the law of division by logarithms, the logarithm of a 
fraction, decimal or otherwise, is negative. 

It is customary to keep the decimal part of a logarithm 
always positive, and to make the characteristic bear the 
negative sign. 

For example, .04324 = ^° 

100000 

.-. log .04323 = log 4324 - log 100000 

log 4324 = 3- 6 359 
log 100000 = 5.0000 



log .04324 = - I.364I = - 2. + .6359. 

This result is usually written, 2.6359, to indicate that 
the 2 alone is negative, while the decimal is positive. It 
is an advantage of uniformity entirely. 



1 1 8 Algebra. 

Again, find log .235. 

lo g • 2 35 ==1 °g-^^-= l°g 2 35 ~ lo S IOO ° 
1000 

lo g 2 35 = 2.3711 

log 1000 = 3.0000 

lo § 2 35 = ~ - 628 9 = "i.37"- 

By an inspection of these results a general rule for pure 
decimals may be stated, thus : 

Find the decimal part of the logarithm from the table, 
ignoring the decimal point. The characteristic is equal to 
a number one greater than the number of zeros following the 
decimal point or is equal to the number representing the 
position of the first significant figure {that is the first one 
not zero) after the decimal point. 

Art. 144. To find the logarithm of a number from a 
table of logarithms. 

Find the log 23.7625. 

Say the table runs to 1000 and gives the log to four 
decimal places. Since the decimal part of the log is 
independent of the decimal point, the point may be placed 
to best advantage ; in this case between 7 and 6, because 
the table gives the logarithms of numbers of three digits. 
Mantissa of 237.625 lies, evidently, between that of 237 
and 238 in the table. If the change of logarithms between 
237 and 238 is uniform, the mantissa of 237.625 should 
be the mantissa of 237 plus .625 of the difference between 
the mantissas of 237 and 238. Thus: 





man 238 = .3766 
man 237 = .3747 




Difference for 
Difference for 


1 = .0019. 

.625 = .0019 X .625 = 
man 237.625 = .3747 + .0012 = 
\ log 23.7625 - 1-3759- 


.0012 4- 
- -3759 



Logarithms. 



119 



The same thing may be represented schematically, 
thus: 

man 238 = .3766 

(man 237.625 = .3759] 



Diff. of one unit = 
Diff. of .625 = 



man 237 



•3747 



Diff. of .0019 corre- 
spondingto one unit 

Diff. of .0012 corre- 
sponding to. 625 



COLOGARITHMS. 

Art. 145. To avoid negative logarithms, where a 
smaller quantity is to be divided by a larger, the logarithm 
of the reciprocal of a number is employed, and is called 
the cologarithm of the number itself. 

For example, find log ^9 § 
562 

log Mi- =log 239 - log 562, 
562 

which would give a negative result. 

-^i- also equals 239 x — - — 
562 562 

... log M9. = i og 239 + i g _1_ 
562 562 

which may be expressed thus : 

log M9_ = log 239 + C olog 562 
562 

log = log 1 — log 562 

562 

log I = 0.0000 
log 562 = 2.7497- 

I 



'- log 

562 



120 Algebra. 

But o may be expressed as 4 — 4 or 5 - 5 or 10 - 10, etc. 
For uniformity we say o = 10 — 10. 

.-. log -^— = (10 - 10) - 2.7497 
562 

= (10 - 2.7497) - 10 = 7.2503 - 10. 



Hence, log = colog 562 = 7.2503 — 10 

562 

.'. log 3" _ log 239 + colog 562 = 
562 

2.3784 + 7.2503 — 10 = 9.6287 — 10 = 1.6287. 
The result may be stated thus : 

TO FIND THE COLOGARITHM OF A NUMBER, FIND ITS LOGA- 
RITHM, SUBTRACT THIS LOGARITHM FROM IO AND WRITE 
- IO AFTER THE REMAINDER. 

The statement of the law for division by logarithms may 
be amended thus : The logarithm of the quotient of two 
numbers equals the logarithm of the dividend plus the 
cologarithm of the divisor. 



To Find a Number From Its Logarithm. 

Art. 146. The number is often called the antiloga- 
rithm. The process of finding a number from its logarithm 
is evidently the reverse of the process for finding the 
logarithm of a number. 

Find the antilog of 3.8764. 

In finding a number from its logarithm the character- 
istic is at first ignored, because only mantissas are given 
in the tables and the characteristics are readily found by 
the simple rules already enunciated. 



: 



Logarithms. 121 

An examination of the table shows no such mantissa as 
8764. The two nearest to it are, 

.8768 = man 753 
and .8762 = man 752. 



The difference .0006 corresponds to difference 1 in 
numbers. 

The mantissa .8764 being between .8768 and .8762 
the number corresponding to it must be between 753 and 
752. which correspond respectively to the mantissas .8768 
and .8762. 

If the change in the mantissa corresponds to the change 
in the numbers, the difference between the smaller man- 
tissa .8762 and .8764 will have the same ratio to the 
difference between .8768 and .8762 as the difference 
between 752 and the number corresponding to .8764 
has to 1, the difference between 752 and 753, i.e., 
.0002 : .0006 : : (x) : 1. 

h = -333 + 



.0006 

.'. man .8764 corresponds to 752.333 + 
.-. log 3.8764 - 75 2 3-33 + 

Since the characteristic of a logarithm is found by taking 
one less than the number of figures in the whole part of 
the number, the pointing off of the whole places in the 
number from the logarithm is the reverse. That is, 
there will be one more place in the whole part of the 
number than there are units in the characteristic. For 
example, the antilog of 1-2345 has two places in whole 
part. The antilog of 3.0642 has four places in whole 
part, etc. 

If the characteristic is negative, the decimal point is 
placed so that the first significant figure in the number 
shall occupy a place after the decimal point of the same 



122 Algebra. 

order as the number of units in the characteristic. For 

example : 

2. 1790 = log 0O151 

1. 7803 = log .603 

4. 6 19 1 = log .000416, etc. 

A solution of a general problem, by logarithms, may 
assist in the comprehension of the process. 
Find value of 



V .008541 2 x 8641 x 4.276^ x .0084 
V (.oo85 4 ) 3 x 182.63* x 82* x~. 4.87. 27T ' 

Log of above expression equals \[2 log .008541 + log 
8641 + ^ log 4.276 + log .0084 + 3 colog .00854 + \ 
colog 182.63 + ^ colog 82 + \ colog 487.27]. 

Man .008541 = man 854.1 
man 855 = .9320 
man 854 = .9315 
diff for 1 = .0005 

difffor.i = .00005 = »oooi (dropping the 5) 
man 854.1 = .9316 [.9315 + .0001] 
log .008451 = 3.9316 ; 2 log .008541 = 5.8632 
or 2 log .008541 = 5.8632 — 10 [adding and subtracting 
10, which does not change value]. 

2 log .008541 = 5.8632 — 10 
log 8641 = 3.9366 
J log 4.276 = .2103 

log .0084 = 7.9243 — 10 

3 colog .00854 = 6.2055 [(10 - 7.7945) - IO ] 

J colog 182.63 = 9.4346 — 10 

J colog 82 = 9.3621 — IO 

\ colog 487.27 = 9-3281 — IO 

Log of original expression = J (52.2647 — 50) 
= J (2.2647) = .5662 
.-. original expression = 2>>^ZZ + = antilog of .5662. 



Logarithms. 123 

It is to be observed that the colog of a decimal, since it 
is the log of the reciprocal of the decimal, is really the log 
of a whole or mixed number, since the reciprocal of a 
decimal must be such a number. Hence the 10 may- 
be subtracted from the characteristic after the colog is 
found without giving a negative quantity as above 

log .00854 = 3". 9315 

3 

3 log .00854 = 7-7945 

[the two carried over the decimal point is positive, hence 
-3X3 + 2= -7]. 

10.0000 — 10 
7-7945 



colog .0085 4 3 = 16.2055 — 10 = 6.2055. 

It must be remembered that the mantissa is always 
positive no matter what the characteristic may be. In the 
above example, since 1 has been borrowed from the 10 
to subtract the .7 from, there are only 9 left from which 
to subtract — 7, hence 9 — (— 7) =9 + 7 = 1 6. 

EXERCISE XXIV. 
Logarithms. 
Find the logarithms of : 

1. 235.6. 4. .00235. 

2. 1.7456. 6. (125.6) 2 . 

3. 1023.5. 6 ( 2 3-67) S - 
Find the antilogarithms of : 

7. 1.301362. 9. 3.673092. 

8. 2.441201. 10. 9.720387-10. 

11. 2.800046. 



124 Algebra. 

Find the cologarithms of : 

12. 216.93. 14. .2765. 

13. .01672. 15. 9929.7. 

Find the value by logarithms of : 
16. 237.95 X .0192. 17. 67.25 -T- 3.2719. 

18. (2.356)* X (77-777) 1 - 



19 ^62.31 X 92086 
•03567 1 

20. // •°3 I 95 2 X 62.932 x -8 3 6 7 8 3 

V 29.312 x (.00261) 4 



2i # 4/ 6.6251 X ^.19672 X .01872 
* ^.51672 X 11. 137 X .09823 

22 (67-025) X (1.06) 1 2 
(1.06) 12 - 1 

23. f V^j -*- yfiT. 

2 4 . Rf) 2 X (f) 3 X ( 3 ^ lf ^ 



25 



' 1/ 2.72 V - 4-6307 m 
•3i7 2 ^124.61 



CHAPTER X. 
INEQUALITIES. 

Article 147. If x — y is positive, x is said to be 
greater than y, written thus, x > y. 

If x — y is negative, x is less than y and this relation 
is represented thus, x < y. 

Art. 148. Two inequalities with the inequality sign 
turned in the same direction are said to be in the same 
sense ; as x > y, 5 > 4, a > b, etc. 

Art. 149. Just as we have equations involving un- 
known quantities, we have also inequalities involving 
unknown quantities. By the solution of equations we get 
values of the unknown to satisfy the equations ; by the 
solution of inequalities we get the greatest or smallest 
value that the unknown may have without violating the 
conditions of inequality. These values are called 
maximum and minimum respectively. 

Art. 150. As equations are subject to certain rules of 
transformation in order that solution may be accomplished, 
so inequalities obey certain laws, which must be deter- 
mined before they can be handled legitimately. They are 
as follows : 

Art. 151. If both terms of an inequality are multiplied 
by a positive quantity, the inequality is unchanged. If 
multiplied by a negative quantity, the sign is reversed. 

Let a > b, then ma > mb, but - ma < - mb. For if 
a > b then a — b = some positive quantity, say c ; that is, 
a — b = c, .'. ma - mb = mc, still a positive quantity. 
.*. ma > mb. 

125 



126 Algebra. 

But - ma — ( — nib) = — mc, a negative quantity, since 
a > b, .'. — ma < — mb. 

Art. 152. If x > y then x m > y"\ 

For x — y = c, a positive quantity. 

Or (of 1 - 1 + x m - 2 v + x m -y. . . xy 7 ' 1 - 1 + y m ) (x - y) 
= £ (x m_1 + x m - 2 y + etc.) which is plainly positive. 

But {m m - 1 + x m ~ 2 y + x m ~ 3 y 2 . . . jc/"- 1 +y») (x-v) 
= x w — /", .". x m — y m = positive quantity, .*. x m > y m . 

Art. 153. \i a i^b (read a is not equal to &) then 
a 2 + b 2 > 2 a6. 

For (a — b) 2 > o (because the square of either a posi- 
tive or negative quantity is positive, hence greater than o) 
that is, a 2 — 2a b + b 2 > o. Add 2 a b to both sides. 

That is, the sum of the squares of two unequal quan- 
tities is always greater than twice their product. 
Example. Find minimum value of x if 






2 x 2 — 8x + 21 > x 2 — 2 x + 37 (collect) 

x 2 — 6x > 16 (add 9 to both sides) 

x 2 — 6x + 9> 25 (extract square root) 

x - 3> 5 .*. x>8. 

Hence, x cannot be as small as 8. 

Example. Find the area of the largest rectangle having 
the perimeter 20 inches. 

Let x = one side, then since the perimeter is 20, 10— x 
= other side 

.*. 10 x — x 2 = area. 

Say, iox - x 2 = y (the area) 

then, x 2 — iox + 25 = 25 - y 

X - 5 = ± ^ 25 - y 

X = 5 ± ^25 - y. 



Inequalities. 1 2 7 

If y is greater than 25, (25 — y) will be negative and 
v 25 — ;y will be imaginary. 

Therefore, y S> 25 (v is w<?/ greater than 25). 

Then 25 is the maximum value of y. 

When 10 x — x 2 = 25 (= y) 

x 2 — iox + 25 = o 

x = 5 the maximum value 
for a side. 10 3: — 5 = 5, 

hence, at the maximum, the rectangle becomes a square, 
5 inches on the side. 

RATIO, VARIATION, AND PROPORTION. 

Art. 154. The ratio of one quantity to another is the 
fraction whose numerator is the first quantity and whose 

denominator the second as — = the ratio of a to b. This 

b 

is often written a : b* 

In such a ratio, a is called the antecedent and b the 
consequent. 

If a and b both change values, but maintain always the 
same ratio, a is said to vary as b, written a °cb. Calling 
m the constant ratio between a and b, this may be written 
a = mb. Clearly one variable quantity may vary as several 
others together ; for instance, a may vary as b, c, d, etc. 
This is expressed thus : 

a oc b.c.d. 

Or a may increase as b decreases (or vice versa); a 
is then said to vary inversely as b, written 

1 
a oc — . 
b 

* The line between the numerator and denominator of a frac- 
tion is probably an evolution from the ratio sign, : . 



128 Algebra. 

Art. 155. There are certain laws governing ratio 
which may be stated as follows : 

If the antecedent is the greater, the ratio is said to 
be of greater inequality. 

A ratio of greater inequality is diminished, and a ratio 
of less inequality is increased by adding any positive 
quantity to both terms. 

Take a > b in the ratio — and let m be any positive 
b 
quantity. 

Add m to both terms, < L±HL 
b + m 

<L±JS. > » or < a - 

b + m b 

According as 

ab + bm > = or < ab + am [clearing of fractions] 

bm > = or < am [subtracting ab from both sides] 

b > = or < a [dividing by m\ 

but a > b by hypothesis, 

hence, - > - ±-^. That is, - was diminished. 

b b + m b 

If a < b, then- < a + m , that is, - was increased. 
b b + m b 

Examples. By Boyle's law of physics if P is the pres- 
sure on a volume V, of gas, then Poc— . A certain gas 

has a volume of 1200 c.c. under a pressure of 1033 g. 
to 1 sq. cm. What is the volume when the pressure is 
1250 g. ? 

Let m be the ratio in the variation P oc — . 

Then P = -or P V - m. «, 






Inequalities. 1 29 

By first condition P = 1033 when V = 1200 
then m = 1033 x 1200 

.-. P = 1033X1200 ( substituting m in Q) 

In second condition P = 1250; 1250 = —23 — £1^? 
24 

_ 1033 x;#0 _ 1033 x 24 _ 68 

EXERCISE XXV. 
Variation. 

1. If y ocx and y = 5 when # = 3, find x when y = 9. 

2. If / 00 — and / =.45 when E = no and /? = 244, 

R 

find £ when / = .48 and R = 254. 

3. If the rate of discharge of water from an orifice 
varies as the square root of the depth, and 1 1 gallons per 
minute are discharged when the height is 49 feet, what 
is the discharge when the height is 77.44 feet? 

4. The distance a body falls, due to gravity, varies as 
the square of the time of fall. If a body falls 257.6 feet 
in 4 seconds, how long will it be in falling 788.9 feet? 

5. The square of the time of revolution of a planet 
about the sun varies as the cube of its distance from it. 
If the distance of the earth is 93,000,000 miles and of 
Saturn 886,000,000 miles, what is Saturn's period about 
the sun ? 

6. A shell 1 foot in diameter weighs ^y 1 ^ as much as it 
would if solid. Find the thickness of the shell, remember- 
ing that the volumes of spheres vary as the cubes of their 
radii or diameters. 



1 30 Algebra. 

7. The penetration of a bullet varies as its momentum, 
or if the mass remains the same, it varies as the velocity. 
If a bullet having a velocity of 187 meters per second 
will penetrate 8.92 cm. into a target, what velocity is 
necessary to penetrate 14.8 cm. ? 

8. The light received upon a surface varies inversely 
as the square of its distance from a source of light. If a 
screen is 25 feet from an incandescent lamp, to what 
distance must it be removed to receive \ as much light ? 

9. The weight of a body on the surface of a material 
sphere varies directly as the mass of the sphere and 
inversely as the square of its radius. If a body on the 
earth's surface weighs 24 lbs., taking the earth's radius as 
3963 miles, what would it weigh on the surface of the 
moon, whose radius is 108 1.5 miles and whose mass is 
•g^th of the earth's mass ? 

10. If the electric resistance of a wire varies directly 
as its length and inversely as the square of its diameter, 
and if a wire 137 cm. long and .038 mm. diameter has a 
resistance of 19.3 ohms, what will be the resistance of 
a wire of the same material 235 cm. long and 1.2 mm. 
diameter ? 

11. Three metal spheres whose radii are 3, 4, and 5 
inches respectively, are melted and cast into one sphere. 
What is the radius of this sphere, the volumes of spheres 
being known to vary as the cube of their radii ? 

PROPORTION. 

Art. 156. A statement of equality between two ratios 
is called a proportion ; thus, 

a: b : : c : d or a : b = c : d or — = — . 

b d 

The first and fourth terms of a ratio are called its 
extremes, the second and third terms are called its 
means. Each ratio is called a couplet. 



Inequalities. 1 3 1 

When a : b : : b : c, b is said to be a mm proportioned 
to a and c, and c ox a are third proportionals to the 
other two. 

A continued proportion is a series of equal ratios, as, 

a : b : : c : d : : m : n : : x 1 y, etc. 

Laws of Proportion. 

Art. 157. Every proportion admits of certain trans- 
formations as follows : 

(a) If four quantities are in proportion, they are also in 
proportion, by alternation, that is the first is to the third 
as the second is to the fourth. To prove, 

If a : b : : c : d then a : c : : b : d. 



Proof, i = L 

b d 



Jf c d c 

Multiplying both sides by -\ or - = - 

c J c d 



(b) Also the product of the extremes equals the product 
of the means. 

For _ = _.-. ad = be (clearing of fractions). 

b d 



(c) They are also in proportion by inversion, that 
is, the second is to the first as the fourth is to the 
third. 



132 Algebra. 

t? a c 

For - = — 

b d 

a b d 

then — = — " 

a c 
[If two quantities are equal their reciprocals are equal.] 

.*. b 1 a : : d 1 c. 

(d) They are also in proportion by composition, that is, 
the sum of the first and second is to either the first or 
the second as the sum of the third and fourth is to either 
the third or fourth. 



That is, a + b 


: a or b : : c 


+ d : c 


or d. 


For 


a - = <L or 
b d 


L = i 

a c 




en <L + 1 = 
b 


- + 1 or 
d 


b - + X 

a 


= i + x 

c 


a + b c + d 


or a + b = 


c + d 


Taddim 



b d a c 

.'. a + b : b : : c + d : d or a + b : a : : c + d : c. 

(e) Prove that they are also in proportion by division, 
that is, a — b : a or b : : c — d : c or d. 

(/) If two proportions have a couplet in each equal, the 
remaining couplets are in proportion. 

If a : b : : c : d and m : n : : c : d then a : b : : m : n. 



For 



i - £- and 2 ■. 

b d n 


c 
~ d 


. a _ m 
b n 




a:b:im:n. 








a 
b 


_ c 
~ d 


and nt = t 

n q 

a _m 
b n 


and c ~ - 
d 


= t 
2 



Inequalities. 133 

If couplets from each proportion form a proportion the 
remaining couplets are in proportion. 
If a : b : : c : d and m : n :: p : q and c : d :: p : q 
then a : b .:: m : n 



For 



(g) In a continued proportion the sum of all the ante- 
cedents is to the sum of all the consequents as any one 
antecedent is to its consequent. 

That is, if a : b : : c : d : : m : n : : x : y, etc. 
a + c + m + x : b + d + 11 + y :: a : b :: c : d, etc. 
For let the common ratio be represented by r then 

— = r or a = br 
b 

— = r or c = dr 



d 
m 

n 



r or m = nr 






x 

— = r or x = yr 

y 

Add ; a + c + m + x = (b + d + n + y) r 

a + c + m +x a c , 

or = r = — = -, etc. 

b + d + n + y b d 

It is readily proved that, if four quantities are in propor- 
tion, any one power (whole or fractional) of these quantities 
forms a proportion. 

Also that if the product of two quantities equals the 
product of two other quantities, two of them may form 
the extremes and two the means of a proportion. 



134 Algebra. 

For if ad = be 



then $L =^- [dividing through by bd] 

b}( "M 



or 



It is to be observed that a, b, c, d, etc. stand for any 
quantities whatever in these proportions. They are by 
no means restricted to monomials. 

EXERCISE XXVI. 
Proportion. 
If a : b : : c : d : 

1. Show that 2 a + b : b : : 2 c + d : d. 

2. That 50 + 36 : 5 ^ - 3 ^ = • 5^ + 3^ : $c - 3d. 

3. Find the mean proportional to 5 and 13! . 

4. Find the fourth proportional to 3^ and 5. 

5. What quantity must be added to each of the 
quantities a, b, c, and d to make them proportional ? 

Find the value of x in the following proportions : 

6. nj : 4i '• : 3i : x - 

_ \m 1 6 m 14.R 

Y, • • • • ^p. 

5 » 7 2? 1 5 » 

8. x : x 2 — 1 : : 15 — 7 x : 8 - 8 ^. 

9 (^ 3 + ;y 3 : x + v : : 7 : 1 
(x 2 — y 2 : x — y : : 5 : 1. 



J + ^3 

10. -- 



V 3-7 



Inequalities. 13$ 

11. Two cars running in opposite directions pass each 
other in 2 seconds ; running in the same direction, the 
faster passes the slower in 30 seconds. What is the ratio 
of their rates ? 

12. What number must be added to each of the numbers 
3, 7, and 13, that the second may become a mean pro- 
portional to the other two. 

13. Show that there is no number that, added to each 
of three consecutive numbers, will make the second a 
mean to the other two. 

14. Spheres are to each other as the cubes of like 
dimensions. What will be the diameter of a ball formed 
from two balls whose diameters are respectively 7 inches 
and 9 inches ? 

15. If a bar is supported at two points, and a weight is 
suspended between these points, the parts of the weight 
borne by the supports are inversely proportional to the 
distances of the weight from the supports. If a bar 18 feet 
long, supported at its ends, carries a weight of 234 lbs. 
4 feet from the end, A, find weight sustained at each end. 

16. The cubes of the planets' distances from the sun 
are to each other as the squares of their periods of revolu- 
tion. Calling the earth's period 1 year, and that of 
Jupiter 12 years, what is Jupiter's distance from the sun 
if the earth's distance is 93,000,000 miles? 



CHAPTER XI. 
PROGRESSIONS. 

Article 158. A series is a number of successive quan- 
tities, each one derived from its predecessor by some fixed 
law. The successive quantities are called terms. If the 
series ends, it is called finite ; if it extends indefinitely, it 
is called infinite. 

Art. 159. There are plainly an unlimited number of 
forms of series. Arithmetical, geometric, and harmonical 
are the only kinds that possess any general importance. 

Arithmetical Series or Progression. 

Art. 160. If each term of a series is derived from 
the preceding term by adding (algebraically) a constant 
quantity, it is known as an arithmetical series or arith- 
metical progression. Such as 1, 3, 5, 7, etc. (adding 
2) ; x + y , x, x - y, x — 2y, etc. (adding - y). 

The general form of this series is, a, a + d, a + 2d 
a + 3d, etc., d being the common difference. 

Art. 161. To find an expression for any term of an 
A. P. (arithmetical progression) in terms of the first 
term, the common difference, and the number of terms, it 
is only necessary to inspect the general form indicated 
above, marking the number of each term. 

(Number of term) 

1st 2d 3d 4th 5th 

a, a + d, a + 2d, a + 3d, a + \d, etc. 

It will be observed that any term is equal to a (first 
term) plus d (difference) taken as many times less one as 

136 



Progressioiis. 137 

the number of this term in the series ; thus the fourth 
term = a + (4— 1) d = a + 3d. 

Calling any desired term /, its number n, first term a, 
and difference d, clearly ; 

I = a + (n — 1) d (x) 

Art. 162. To find the sum of any number of terms : 
With the same notation, in addition calling the sum s, and 
remembering that if, starting with the last term, the 
common difference be subtracted successively, the series is 
the same (but in reverse order) as if we started with the 
first term and added the difference, then ; 

s = a + (a + d) + (a + 2d) . . . (/ - d) + /, 
s = I + (I - d) + (I - 2d) . . . (a + d) + a 

Add, 2 s = (a + /) + (a + /) + (a + /) + (a + /) . . . 
(a+ /) + (a + /). 

If there are n terms in the series, evidently there will be 
n (a + l)'s, that is, 2 5 = (a + I) + (a + I) + (a + /) 
-f- repeated n times ; or, 2 s = n (a + /) 

j - 5- (a + /)\ . . . . (y) 

2 

(x) and (y) are the fundamental relations between parts 
of an arithmetical progression. From them and their 
combinations, if any three of the quantities </, d, », s, and 
/ are given, the other two are readily found. For 
example : 

In an arithmetical progression, given 

d = 7, n — 12, 5 = 594. 

Substituting in (v), / = a + (12 - 1) 7 = a+ 77 
or *-<* = 77 (0 



138 Algebra. 



Substituting in (y), 594 = — (a + /) = 6 (a + I) 

2 

or a + / = 99 (2) 

Adding (1) and (2), 2/ = 176, / = 88 
Subtracting (1) from (2) 

2a = 22, a = 11. 

Arithmetical Mean. 

Art. 163. The arithmetical mean between two quan- 
tities has a wide field of application in practical mathe- 
matics. It is the quantity which, placed between two 
other quantities, forms with them an arithmetical progres- 
sion. 

If a and I are any two quantities, and m stands for their 
arithmetical mean ; by definition, a, m, I is an arithmeti- 
cal progression ; 

hence m — a = I — m (since there is a constant difference) 

or 2m = a + / 

m = a + I 



Art. 164. It is sometimes necessary to insert several 
arithmetical means between two quantities, and. here for- 
mulae (x) and (y) assist. For example : 

Insert 4 arithmetical means between 3 and 18. 

If m v m 2 , m 3 , m v stand for the means, the series will be, 
3, m v ?n 2 , m v m 4 , 18, hence a = 3, n = 6, / = 18, to find d. 

Substituting in (x), 18 = 3 + (6 — 1) d 
whence, 5^ = 15; d = 3 

means 



3, I 6, 9, 12, 15,1 18 is the series. 



Progressions. 1 39 

Art. 165. It is convenient to represent unknown quan- 
tities, when they are in arithmetical progression, by the 
following series, the first one when the number of 
unknowns is odd ; the second when it is even ; etc. 

x — 2v, x — y, x, x + y, x + 2y, etc. 
x — 3y, x — y, x + y, x + 3y, etc. 

As an illustration : 

The sum of three numbers in arithmetical progression 
is 33, and the sum of their squares is 461. Find the 
numbers. 

Let x — y, x, and x + y represent the numbers, 
then (x - y) + x + (x +y) = 33 . . . (1) 

(x - y) 2 + x 2 + (x + y) 2 = 461 . . . (2) 

From (1) the wisdom of the above notation is evident, 
for it reduces to, 3 x = 33, x — 11 
hence, from (2), (11 - y) 2 + (n) 2 + (11 + y) 2 = 461 

y = 7- 

EXERCISE XXVII. 
Arithmetical Series. 

1. Find the 8th term of the series 3, 8, 13. . . . 

2. Find the 10th term of 2j, if, ij. . . . 

3. Find the 9th term of 3, 2 J, if. . . . 

Find the sum of : 

4. 1+3 + 5 + 7. ..to 15 terms. 
6. —3+1 + 5. ..to 10 terms. 

6. 4 + 1 + ^. ... to 20 terms. 

7. x + (3 v — 2 y) + (5 x — 4 v) ... to 8 terms. 



x - 1 + x - 3 

X X 

9. Insert 6 means between 9 and 177. 



8. h ■ £ + ... to 12 terms. 

x x 



I4 o Algebra. 

10. Given a = '3 \, I = 64, n = 82. Find ^ and 5. 

11. Given / = 105, » = i6 : s = 840. Find a and d. 

Find parts not given in the following : 

12. d = 5, / = 77, 5 = 623. 

13. 5 = 143k a = I, » = 20. 

14. n = 20, a = 5, d = 2§. 

15. a = 200, / = 88, s = 2160. 

16. d = 4, n = 14, 5 = 812. 

17. How many terms of the series — 5, — 2, +1 . . . 
must be taken to sum 63 ? 

18. The first term of an arithmetical progression is 5, 
the third term is 17. Find the sum of 8 terms. 

19. How many terms of the series 2, 5, 8 . . . must be 
taken, that the sum of the first half may be to the sum of 
the second half as 8:23 ? 

20. Starting from a mark, 30 stones are placed at 
intervals of two feet. If, starting at the mark, the stones 
are collected one by one and carried back each time to 
the mark, how far will the collector walk ? 

21. The three sides of a right angled triangle, whose 
area is 54 square rods, are in arithmetical progression. 
Find the sides. 

22. If a falling body descends 16.1 feet the first second, 
48.3 feet the second second, 80.5 feet the third second, 
how far will it fall in one minute ? 

23. A man was given his choice of wages, either $1.00 
per day, or 3 cents the first day, 6 cents the second, 
9 cents the third, his wage increasing 3 cents each day. 
He chose the former. Did he win or lose (in 30 days) and 
how much ? 

24. What value of m will make the arithmetical mean 
between a? and a% equal to 6 ? 



Progress ions. 1 4 1 



Geometrical Progression. 

Art. 166. A geometrical progression is a series of 
quantities so related to one another that each bears a 
constant ratio to the preceding. Thus: 2, 4, 8, 16, etc., 
or in general, a, ar, ar 2 , ar 3 , . . . etc. 



Value of Any Term. 

Art. 167. Let a represent the first term, /, the last 
term, r, the ratio, n, the number of terms, and s, the sum 
of the series. 

Then a G.P. (geometrical progression) is represented in 
general by 

(Number of term) 1st 2d 3d 4th 5th 

a, ar, ar 2 ,ar 3 , ar 4 , etc., 

the numbers indicating the number of the term. 

It will be observed that any term is the product of the 
first term, a, by r, raised to a power, which is one less than 
the number of the term in the series. Hence, if n repre- 
sent the number of any term, /, in the series, this term 
will be 

/ = ar n ~ l ...... (1) 

which is one of the fundamental relation equations for 
geometrical progressions. 

Sum of Any Number of Terms. 

Art. 168. According to the notation above, evidently, 
s = a + air -f- a/r 2 + a/r 3 +a'r 4 + . . . ar"' 2 + ar*- 1 (or /) 

multiplying by r / / ^^ 

rs - a/r + afr 2 + a[r 3 + a/r 4 . . . a.r"-^ + ar n or (//•) 



142 Algebra. 

Subtracting, s - r s = a - ar", (or a — rl) 

or s (r - 1) = a (r n - 1), (or rl - a) 

a (r n — 1) rl - a , N 

s = — ^ ' or . . (2) 

r — 1 r - i 

Art. 169. By these formulae or by combinations of 
them, as in the case of arithmetical progressions, any 
two of the quantities, a, d, n, /, and s, may be found when 
the other three are given. 

Example. Given a = 5, w = 3, 5 = 285. 



From (1) 


1= 5 r 2 


From (2) 


285- 5(' 3 -0 
r - 1 




3 

57 = r ~ =r 2 + r+ 1 
r — 1 




r 2 + r = 56 

f 2 + r + 1 = 2|5 



r = 7 or — 8 

whence, from (1), / = 5 (y) 2 or 5 ( — 8) 2 =245 or 320. 
The series is, then, either 5, 35, 245 or 5, -40, 320. 

Geometrical Mean. 

Art. 170. A geometrical mean between two quantities 
is a quantity which bears to the first quantity the same 
ratio that the second quantity bears to it ; that is, it forms 
with the two quantities a geometrical progression. 

If x stand for the geometrical mean between a and b, 
then by definition : 

— = — or x 2 = ab x= \/~ab 
a x 

That is, a geometrical mean between two quantities is 
equal to the square root of their product. 



Progressions. 143 

For example, the geometrical mean between 16 and 25 



is ^16 x 25 =20; between a + 2 b H and a is 

a 



VA 



lr 



a (a +26+-)= ^a 2 + 2 a& + b 2 = a + b, etc. 



a 

Art. 171. If several geometrical means are inserted 
between two quantities, each of these means is a mean 
between the two means on either side of it. Hence, to 
insert any number of means between two quantities, it is 
necessary to construct the series, hence to find r. 

Example. Insert 5 geometrical means between 2 and 
1458, calling the means G v G v G r G r and G 5 , the series is 
2, G v G 2 , G v G 4 , G 5 , 1458 and n = 7. 

By (1) 1458 = 2 r 6 

or 76=729 

r = V729 -3. 

Hence, the series, 2, 6, 18, 54, 161, 486, 1458. 

Rule: Taking the two quantities as a and t and n 
equal to two more than the number of means, use 
formula (1). 

Infinite Series. 

Art. 172. If the number of terms is unlimited, the 
geometrical progression is called an infinite series, other- 
wise it is finite. 

Formula (2), s = -* '-, may be put in the form, 

r - 1 

ar n a a ar" 



r — 1 r — 1 1 — r 1 — r 

If r is a fraction and n is large enough, the value of r n 
may become insignificant, since each increasing power of 



144 Algebra. 

fraction is less than the preceding one, for by definition 
a power of a quantity is the product of a quantity by itself 
a certain number of times, and if the quantity is a fraction, 
the product is each time multiplied by a fraction, which 
must decrease its value. 
If then n is infinite r n = o, 

o hence, 5 = — - — . (3) 



1 — r 1 —r 1 — r 

which is the formula for the sum of an infinite geometrical 
progression. 

Example. Find the sum of the infinite geometrical 
progression 1, J, J, J. . . . 

Here, 



a 


= 


1, 


r 


= 


1 
2 


, n 


= 


00 








s 






a 






1 






I 




2. 




1 


— 


r 




1 - 


1 




1 





Art. 173, A common application of the formula for 
infinite geometrical progression, is the expression of a 
recurring decimal in terms of a simple fraction. 

Example. Evaluate .124545. . . . 

This is equivalent to the series ■ 

.12 + .0045 + .000045 + • • • 

or T Vo + T o 4 ifW + Tooeoo +■..••; t0 innnit y> 

as the figures 45 are repeated indefinitely in succeeding 
decimal orders. 

Starting with j-^qq the rest of the expression is plainly 
an infinite geometrical progression in which r = t ^q and 
a = toVoo-- 

• c _ a T0 4 oW _ ToVoO _ 4 5 

• ' s " 7T7 " t _ _L_ " ^vr ~ ™™' 

1 ' x 100 100 

Then .124545 . . . = T V 2 <> + 9M0 = if 88 = tt¥o- 
The recurring digits in such a decimal are usually indi- 
cated by a dot placed over them. 

Thus, in the example above, .124545 . . . = .1245. 



Progressions. 145 



Harmonical Progression. 

Art. 174. A harmonical progression (H.P.) is a series 
of numbers whose reciprocals, in the same order, form 
an arithmetical progression. 

Thus, i, |, \, \ is an harmonical progression, since 3, 
5, 7, 9 is an arithmetical progression. 

Again, f, 4, — f, —\\ is an harmonical progression, 
since §, \, — f, — f| is an arithmetical progression. 

Hence, to solve an harmonical progression invert its 
terms and apply the formulas for an arithmetical pro- 
gression, then reinvert. 

EXERCISE XXVIII. 
Geometrical Progression. 

1. Find the 10th term of 3, 6, 12. . . . 

2. Find the 9th term of 6-*-, 2^, §f. . . . 

3. Find the 7th term of 32, —16,8. . . . 

4. Find the 6th term of if, 2§, d^L. . . . 

Find the sum of : 

5 - 5 + (-3) + T t • • • to 9 terms. 

6 - l + ( -I) + £s • • • t0 IO terms. 

7 - i + h + I • • • to infinity. 

8 - A + tV + 1 + • • • to 8 terms. 

Find parts not given in following : 
9. a = 36, / = 2], n = 5. 

10. / = 1296, r = 6, s = 1555. 

11. r = 2, n = j, s = 635. 

12. a = -§, a = 7, r = -J. 

13. a = 1, I = 81, r = 3. 



146 Algebra. 

14. Insert 3 geometrical means between 17 and 4352. 

15. Insert 6 geometrical means between 5 and — 640. 

16. The fifth term of a geometrical progression is 48 
and r = 2. Find first term. 

17. Four numbers are in geometrical progression. The 
sum of the first and fourth is 195, and the sum of the 
second and third is 60. Find the numbers. 

18. The sum of the first 8 terms of a geometrical pro- 
gression is 17 times the sum of the first 4 terms. Find 
the series. 

19. Find the value of the recurring decimal 

3.17272. . . . 

20. Find the value of the recurring decimal 

•I53I53- • • • 

21. A blacksmith proposes to shoe a horse for $1.60 or 
to take for his work, i cent for the first 4 nails, 2 cents 
for the next four, 4 cents for the next four, and so on. If 
he used 8 nails to each of the four shoes, which proposition 
was the better for him ? 

22. A "letter chain " is started thus, for a memorial 
fund : three letters are sent out with a request for 10 cents, 
and each receiver is asked to send out three letters con- 
taining the same requests. This process is repeated 30 
times. How much will be realized for the fund ? 

23. If $100 be placed in a savings bank, where the 
amount increases 4 per cent each year, how much will be 
to the depositor's credit at the end of 20 years if no money 
is withdrawn ? 



CHAPTER XII. 
INTEREST AND ANNUITIES. 

Interest. 

Article 175. 

Definition. Interest is the earnings of money when 
loaned or invested. 

Definition. The principal is the sum thus put to use. 

Definition. The ratio of the earnings for one year to 
the principal is called the rate oj interest, or simply the 
rate. 

Definition. The amount is the sum of principal and 
interest. 

Definition. When the interest itself earns interest at 
stated intervals, it is said to be compounded. Such interest 
is, hence, called compound interest. 

Simple Interest. 

Art. 176. Let p = principal ; r = rate; n = time; 
A = amount ; / = interest. Then by arithmetic, / = Pm 
and A=P + I = P + Pm = P (I + rn). 

Compound Interest. 

Art. 177. By definition, if the interest is payable 
annually, 

A = p (1 + f) = PR [letting 1 + r - R] ; end of 1st year. 
A, = PRr + PR = PR (1 + r) = PR 2 ; end of 2d year. 
A\ = PR 2 r + PR 2 = PR 2 (1 + r) = PR 3 ; end of 3d year. 
147 



148 Algebra. 

A comparison of the exponent of R with the number of 
years will enable us to express the amount for any number 
of years, say n, thus ; 

A n - PR n . 

The subscripts for A indicate the number of years for 
which the amount (^4) stands. 

Art. 178. Frequently the interest is compounded 
semi-annually or quarterly, as in savings accounts. Then 
again by definition, if the interest is semi-annual, 

A h = \Pr + P = P (1 + -), for 1st half year. 



h 



A ± - iP(i + r -) r + P (1 + - P (1 + Q 2 , for 2 d 

half year. 



A* = + P 1 1 + - ) r + P [ 1 + - ] - P [ 1 + -V ; for 



3d half year. 

Then by analogy for n years or 2 n half years 



L ,» 



If the interest is quarterly, exactly similar process gives 
the formula 

/ Y \ i n 

A, 



-(' + 5J 



Annuities. 

Art. 179. An annuity is a fixed amount of money to 
be paid or set aside annually or at stated regular intervals. 

If these amounts are allowed to accumulate at com- 
pound interest, the annuities constitute a sinking fund, 
which is usually a provision for eventually liquidating an 
indebtedness of some institution. 



Interest and Annuities. 149 

Art. 180. Let S be the annuity ; R, the amount of one 
dollar for one year at the rate, r ; n, the number of years, 
and A, the final amount of the sinking fund, at any num- 
ber of years. Then by definition, 

A 1 = S, at the end of 1 year 

A 2 = S + SR, at the end of 2 years. 

A 3 = S + SR + SR 2 , at the end of 3 years. 

A, = S + SR + SR 2 + SR 3 , at the end of 4 years. 

And by easy analogy, 

A n = S +XR y&R 2 jySR 3 + . .^Sli 71 - 1 . (1) 
.'. A n R - SR + ^+Sf^+. . . SP^+SR n . (2) 

(multiplying (1) by Z?) 

Subtract (1) from (2), 4 B # - 4„ = 57?" - 6* 
or A n (R - 1) = S(£" - 1) 

.*. /I, = 5 (fl* - 1) 
fl - 1 

Art. 181. It is usually necessary in the establishment 
of a sinking fund to estimate the amount required as 
annuity, or the number of years with a given annuity, to 
meet the obligation assumed. 

Let P = the amount of debt. 

R = the amount of $1 at rate r, for 1 year. 
n = time. 

Then if the debt and its, accumulated interest are to be 
balanced by the annuity and its accumulations, 

PR n (the debt and compound interest for n years) 
= ^-^ - (the annuity S, and its accumulation at the 

same rate). Solving this equation for S, 



150 Algebra. 

By the use of logarithms n may be found from (a) if S 
is given or vice versa. 

Example. What annuity will satisfy a debt of $4600 
in ten years, money being worth 5 per cent? 

(a) becomes, 5 = 46°o (i-QS) 10 (.05) . ^3° (i-QS) 10 
(1.05)" - 1 (i.o S )» - 1 

log (1.05) 10 = .021189 x 10 = .211890 
log 230 = 2.361728 

log 230 (1.05) 10 = 2.573618 

(1.05) 10 = anti log .211890 = 1.6289 
(1.05) 10 - 1 = 1.6289 - 1 = .6289 

log S = log 230(1. 05) 10 + colog .6289 = j 2 '2oi 4 i8 

2.775036 



EXERCISE XXIX. 
Interest and Annuities. 

1. Find what $i would amount to at 6 per cent, 
compounded annually in 20 years. 

2. What sum will in 8 years at 5 per cent com- 
pounded annually, amount to $1327.67 ? 

3. A certain principal will in 7 years at a certain rate, 
simple interest, amount to $1136, and in 10 years to 
$1280. Find principal and rate. 

4. In what time will $960 at 6 per cent, annually com- 
pounded, amount to $1190.48 ? 

5. Find the difference between the amount of $1200 
when compounded annually at 6 per cent and when com- 
pounded quarterly at same rate. 



Interest and Annuities. 151 

6. Find the present worth of $7500, due in 6 years, if 
money is worth 5 per cent compounded annually. That 
is, find the principal that will amount to $7500 in 6 years, 
at 5 per cent compound interest. 

7. A church borrows $1000, and renews its note every 
six months at an increase of 10 per cent. How long will 
it take the note to reach $4594.97 ? 

8. How long will it take a sum of money to double 
itself at 5 \ per cent compound interest, compounded 
annually ? 

9. An institution borrows $10,000. What amount 
must it set aside yearly to pay the debt in 15 years, money 
being worth 5 per cent ? 

10. What annual premium must an insurance company 
charge that it may pay a policy holder $1000 at the end 
of 15 years, and still make $200, if money is worth 5 per 
cent ? 



PLANE TRIGONOMETRY. 



•53 



PLANE TRIGONOMETRY. 

PART I. 

THE RELATION BETWEEN ANGLE AND LINE. 

Article i. Trigonometry is a branch of mathematics 
which is concerned with the estimation of lengths, areas, 
and volumes, by using the relation between angle and line 
as well as that between line and line. 

Geometry affords no general relation between angles 
and lines ; it offers no method of comparison between an 
angle expressed in degrees and a line expressed in feet 
or inches ; trigonometry, however, enables us to make 
such comparisons. 

Suppose, for instance, we wished to measure the dis- 




Fig. x. 

tance from a point A to an inaccessible object P; further, 
imagine a wood W so situated that B would be the first 
point to the right of A from which an unobstructed view 
of P could be obtained. Now, we might lay off a known 
length AB (called a base line) and with a transit, measure 
the angles BAP and A BP ; we could then calculate the 

i5S 



i 5 6 



Plane Trigonometry. 



angle APB, Now by geometry we could obtain no in- 
formation from these data regarding the length of AP, 
but trigonometry enables us to involve the known angles 
and side AB in. calculation, so that^lP can be determined 
in linear units. 

Art. 2. The necessity of being able to combine in 
calculations, angles expressed in degrees, minutes, and 
seconds, with lines expressed in linear units, led to the 
invention of six new expressions called functions of angles. 
In general, one quantity is said to be a junction of another 
when it depends upon it for its value ; now the six trigono- 
metrical functions are usually expressed as ratios between 
the sides of a right triangle, in which the angle concerned 
occurs, and although they are entirely linear, depend 
directly upon the angle for their values. 

Being linear, they readily combine with the sides, and 
depending as they do upon the angle for their value, the 
latter becomes a useful element in calculation. 



Trigonometric Ratios. 
Art. 3. We will now take an angle KMH (see Fig. 2) 




having the special size 32 15' '; upon one of the sides MH 
take a point P and draw PN perpendicular to MK. We 
thus obtain a right-angled triangle PNM, with the right 



The Relation Between Angle and Line. 157 



angle at N. To find the ratio of NP to MP we may 
proceed as follows : taking any unit of length, say fa 
inch, we find by careful measurement, NP = 56.5, MP = 

106; hence the ratio = 5 '5 = 0.5330. 

MP 106 

Constructing in like manner any other perpendicular 

N'P' and taking the millimeter, suppose, as our linear 

N f P f 
unit, we find N'P'= t>°-75 and AfP' = 57.7 ; hence 



MP' 



VP N'P' 



; now our 



0.5326. 

We know from geometry that — 

h ' MP MP' 

actual measurements agree very approximately, and the 
mean of the results gives 0.5328. It is possible mathe- 
matically to calculate the ratio of the side opposite to the 
hypotenuse for a given angle to any degree of accuracy ; 
for the special case of 32 15' we find 



length of opposite side 
length of hypotenuse 



o.5336, 



a quantity to which our rough measurements approximate, 
and which is independent of the unit of linear measure- 
ment. This quantity is called the sine of 32 15'. 

50' and draw- 

N' 



Now taking the angle SMK which is 2 




ing any two perpendiculars NT and N'T', we obtain in 
terms of fa inches, NT = 27.5, and MN = 74.0, giving 



i5« 



Plane Trigonometry. 



NT _ 27.5 
MN /4.0 

and J/X 



0.3716; again in millimeters, N'T f = 19-75, 

ATT'- 19-75 
53- 



53-25>g 1 vmg 



710. The 



MN' 53.25 
mean of these results is 0.3713, and the accurate value cal 

1 , , , n 1 . side opposite „ 

culated to nve places is *-*- — — = 0.37 191. Hence 

hypotenuse 

we say the sine of 21 50' = 0.37 191. It is easy to see 

that the sine, although constant in value for any given 

angle, is dependent directly upon its size ; hence we say 

the sine of an angle is a function of it. 

Art. 4. The six trigonometric functions are as fol- 
lows : — a right triangle MNR being constructed with 
Z.M as one acute angle. 

The sine of M is the ratio of the opposite side to the 
RX 



hypotenuse, or 



MN 



The cosine of M is the ratio of the adjacent side to the 

hypotenuse, or . 

* F MN 




The tangent of N is the ratio of the opposite side to 
the adjacent side, or — — • 

The cotangent of M is the ratio of the adjacent side to 

u. ■* -a MR 
the opposite side, or -. 

FF RN 



The Relation Between Angle and Line, i^c) 

The secant of M is the ratio of the hypotenuse to the 

adjacent side, or . 

J MR 

The cosecant of M is the ratio of the hypotenuse to the 

opposite side, or . 

1F RN 

In the right triangle MNR the capital letters repre- 
sent angles and the corresponding small letters represent 
the opposite sides, R being the right angle. 



Then 



ite sine of M . 


. . sin M, 




cosine of M . 


. . cos if. 




tangent of M . 
cotangent of M . 
secant of M . 


. . tan M, 
. . cot if. 
. . sec M. 




cosecant of M . 


. . esc M. 




sin if = — {a), 
r 


cot M - — 
m 


(d). 


cosM =— (b). 
r 


sec if = — 

n 


(e). 


tanM =— (c). 

n 


CSC if = ~ 

fff 


(/)• 



Art. 5. It will be observed by comparing (a) with 
(/), (b) with (e), and (c) with (d), that esc if = —^ — , 

sec if = — - — , and cot N = ; hence if the three 

cos M tan if 

functions, sine, cosine, and tangent, are known, the others 

may be easily found by taking the reciprocals of these. 

The functions most commonly used are the sine, cosine, 
tangent, and cotangent. 

Art. 6. By producing the two sides including the 
angle if, and drawing perpendiculars from various points 



i6o 



Plane Trigonometry, 



of NM produced, to NR produced, it is evident, by simi- 
lar triangles, that so long as M remains the same, the 
ratios do not change, although both terms of any ratio 




alter in value ; thus, Produce MN to and MR to P 
draw perpendiculars to MP from MO. Then sin M = 
RN • __„_.,_ ^ TT ,. _.-__ „ KL 



MN 
sine M 



in triangle MNR ; sine M 
OP 



in MOP, but 



ML 
NR KL 



MN ML 



in ITXZ,, and 

OP , . . 
by simi- 



1/0 
lar triangles. 

On the other hand, if if changes, the ratios change. 
For, let M increase, and suppose the hypotenuse to re- 
main the same, then the side opposite increases, and 
hence the ratio of opposite side to hypotenuse increases, 
etc. 

When the angle M is zero, since the side M decreases 
with the angle, and ultimately vanishes, the sine of zero 

In the case of the 



degrees takes the form — = - 
r r 



o. 



tangent we have 



o. Thus we find sin o° = o, tan o° 



= o. We further note that, with a decreasing angle, the 
hypotenuse r becomes more and more nearly equal to n^ 



The Relation Between Angle and Line. 161 



and when the angle M is finally zero, 
the cosine and secant each become — ■ 



n, and hence 
In the case 



of the cotangent and cosecant, since it is the side m that 
now decreases as the angle becomes less, when M van- 
ishes we have for these functions the ratios — and — , each 

o o 

of which equals infinity. We thus obtain the values sin o° 
= o, cos o° = i, tan o° = o, cot o° = oo , sec o° = i, 
cosec o° = oo . By a similar process of reasoning we find 

sin 90 

,0 _ 



1, cos 90' 



oc , cot 90 = o, sec 



o, tan 9<d v 
90^ = 00, cosec 90" = 1. 

A comparison of the results, which are of considerable 
importance and should be carefully noted, shows that the 
values of the sine and cosine of an angle from zero to 90 
cannot be greater than unity, while the values of the other 
functions vary between zero to infinity. 

Art. 7. Since 6o° and 45 are angles of equilateral 
or isosceles triangles, geometry enables us to find their 
functions very easily, and also those of 30 , which is half 
the angle of an equilateral triangle. 



To Find the Functions of 45 . 

Art. 8. Let xyz be any 
isosceles right triangle, y being 
the right angle. Then Z.v = Zs 
= 45 , and as the sides xy and 
yz can be of any length, we 
will put 



xy = yz 



a. 




Now 



and 



xz' 



xz 



= x y + yz . 
- a 2 + <r - 
= a V2, 



2 n 



Fig. 6 



1 62 Plane Trigonometry. 

Thus we get. 



sin 


a 

45° = -7=- - 
aw 2 


1 
V2 


2 


cos 


a 
45° - -/=■ = 

flV2 


1 
Vi" 


_ V2~ 
2 


tan 


45°=^ = - = 
a 1 


= 1 




cot 


a 1 


= 1 




sec 


45 = = 


V7 

1 


- V7 


cose 


aV2 
C45 = = 


V2 


= vT. 



It will be noticed that, 



sin 45" 
tan 45 



cos 45 u = 



V2 



0.7071 + 



cot 45 = 1 



sec 45 = cos 45 => V2 = 1.4142 + 

To Find the Functions of 60 ° and 30 . 
Art. 9. Let ABC be an equilateral triangle. Draw 

B 

A* 




the perpendicular BD from B to AC; then Z.BAD 
= 6o° and Z.ABD = 30 , also Z,4£>£ = 90 . 



The Relation Between Angle and Line. 163 

Now AD = \ AB ; let x be the length of AB, then 

AD = -; 

2 

and since DB 2 + IP 2 = AB 2 



we get Pi*' 



© ; 



.\ £>£* = * 2 - - = ^* 2 
4 4 

and ££> = -^ *. 

2 

Hence sin 6o° = — = ^ cot 6o°= -L = -^2- 

x 2 ^3 3 

cos 6o°= t? = 1 sec 6o° = 2 

— ^x __ 

tan 6o° = — = VJ cos 6o° = -^= = 2x/3 . 

J x ^3 3 



If we now note that DB is the side adjacent to the 
angle ABD = 30 and .4.D the side opposite to it, we 
further obtain 



1 

sin 30°= ^ = - cot 30 = V3 



o _ vjc _ 1 



# 2 
V3 



cos 3 o° = -1- L ^3_ sec 30 o = ^3 
* 2 3 

tan 30 = _i*- -= - - 1 ^5- cos 30 = 2. 

^3 x V 3 3 



164 Plane Trigonometry. 

Comparing the above two sets of values, we find, 



ci'n f\oP — pnc '»i~i < ^ — ^ 


= 0.8660 


oi.il UU — LU3 O — 

2 


cos 6o° = sin 30 = — 

2 


= 0.5000 


tan 6o° = cot 30 = V3 


= 1.7321 


rot 6n° — tan m° — 3 


= o-5774 


L,UL UU — Ld.il s LJ — 

3 


sec 6o° = cos 30 = 2 


= 2.0000 



cos 6o° = sec 3o c 



V 3 



= I -i547 



A simple reference to the figure has shown that the func- 
tions of 6o° are the co-functions of 30 , and conversely; 
this would clearly be true for any right triangle and any 
acute angles. 

Since the latter are complementary, the significance of 
the prefix " Co " is revealed. It is simply an abbreviation 
for complementary . That is, a function of any angle is the 
corresponding co-function of its complementary angle, and 
conversely. 

Since angles of 45 °, 30 , and 6o° are of frequent occur- 





Fig. 6a. 



Fig. 7a. 



rence, the functions of these angles are important. It 
will be found easy to recall the numerical results of Arti- 
cles 8 and 9 if we assign special values to the sides of the 
triangles in Figs. 6 and 7 ; this we are at liberty to do, 



The Relation Between Angle and Line. 165 

as the values of the functions are independent of the unit 
of length chosen. In Article 8, putting a = 1 we have 
xz = V2 ; again, in Article 9, if x = 2, AD = 1, and BD 
= V^. We thus get Figs. 6a and ja, from which it is 
easy to obtain the numerical value of any function of 45 °, 
30 , or 6o° when required ; for instance, a glance at Fig. 

7a shows that sin 60 = — ^- , sin 30 = _ , etc. 
2 2 

The functions of angles in general cannot be found 
thus simply by geometry, but are estimated from series, 
into which the six functions have been developed by 
methods of higher mathematical analysis. These values 
and their logarithms are set down in tables, which record 
them to single minutes, and sometimes to smaller parts of 
a degree. By a method called interpolation, explained in 
connection with these tables, the functions of any angle, 
or the logarithms of these functions, may be found. 

Art. 10. As already explained, the six functions can 
be grouped into three pairs of reciprocals (see § 5) ; 
thus in the triangle ABC, 

esc A sin A = 1 ) 
sec A cos/1 =1 Mi) 
tan A cot A = 1 ) 

The question now arises, How can we find further rela- 
tionships among the functions ? We have a right-angled 
triangle to work upon, and have already used the comple- 
mentary property of the acute angles to discover that the 
functions of the one are the co-functions of the other. 
Now what further property of the right triangle remains 
to be investigated ? Can we obtain new relations among 
the functions from the equation a 2 + b 2 = c 2 ? Clearly we 
can by division ; that is, if we take each term separately 
and divide by it we shall obtain three new equations in- 
volving the squares of the functions. 



1 66 Plane Trigonometry. 



Taking 






a 2 + b 2 = c 2 


a 2 + b 2 = c 2 


a 2 + b 2 - c 2 


Dividing by c 2 
a 2 , b 2 

~2 + ~2 = * 

c 2 c 2 


Dividing by 6 2 
a 2 c 2 
b 2+I = b 2 


Dividing by a 2 

1 + *. 2 = e i 

a 2 a 2 


e-)"-C-)'- 


(r)"— 0' 


■ + ©"-(;-)" 


sin 2 ^4 + cos 2 yl = i 


tan 2 A + 1 =sec 2 ^4 


1 + cot 2 ^4 = esc 2 A 


We thus obtain 


= 1 ; sin A = Vi 




(2) sin 2 A + cos 2 A 


— cos ^4 ; cos A 




= y/i — sin" .4 





(3) sec 2 A = 1 + tan 2 ,4 ; sec ^4 = Vi + tan yl ; tan A 

= ^sec 2 ^4 - 1 

(4) cosec 2 .4 = i + cot 2 .4 ; cosec .4 = Vi + cot 2 A ; cot A 

= Vcsc 2 A — 1 

It will be noticed that since sin A = — and cos A = -- .\ 

c c 

S -™A . <L . £ _ £ = tan 4. 

COS /I £ 6 6 

We thus find two important relations : 

. sin A a 4- a cos A 

tan ^4 = ■ and cot A = . 

cos A sin A 

EXERCISE I. 
Functions as Ratios. 

With the usual notation, large letters representing the 
angles of a right triangle, and the corresponding small 
letters the sides opposite, find the functions of B, A being 
the right angle, when, 

1. a = 17, b = 8, c = 15. 

2. a = 15, b = 9, c = 12. 

T 

3. a = 12, b = 10, c = 7— ♦ 

2 





The Relation 


Between Angle and Line. 


4. 


a = 35.7, b - 


31.5, c = 16.8. 


5. 


a - 5-55, 6 = 


4-44, f = 3-33- 


6. 


a = 17.85, b = 


8-4, c = 15.75. 


7. 


a = 25.8, b = 


15.48. 


8. 


6 - 


158.1, c = 74.4. 


9. 


a = 61, b = 


11. 


10. 


a = 1. 


e-h 


11. 


c = 2b. 




12. 


c + b = if j. 




13. 


a — c = ^6. 





167 



Find the other sides of the right triangle, if, 



14. 


sin B = T 4 T and a = 22. 


15. 


cos B = % and c = 6 . 


16. 


tan £ = l± and a = 61. 


17. 


sec C = | and b = 9.6 


18. 


Two straight roads make 



angle 01 40^; a man 
walks down one road 1^ miles, and then crosses over to 
the second road in a straight line, meeting it at right 
angles. How far from the starting point will he be in a 
direct line, if sin 40 = .6428 ? 

19. The grade of a railroad track is about 10^, that is, 
it makes an angle of 6° with the horizontal. What weight 
can a locomotive pull up the grade, if it can haul 500 tons 
on the level? sin 6° = .105. 

Find the other functions, if, 



20. 


sin x = }. 


21. 


cos 50 = .766, 


22. 


tan 45 = 1. 


23. 


sin 90 = 1. 


24. 


cot 6o° = J v^ 


25. 


2 
sec 30 = --=■ 

v 3 



i68 Plane Trigonometry. 

Find the angle A , given ; 
26. sin A = cos 2 A. 

Solution. To compare the two sides of this equation, 
it is necessary to express both in the same function, and 
since sin A = cos (90 — A) [or cos 2 A = sin (90 — 2 A)], 
cos (90 — A) = cos 2 A [or sin A = sin (90 — 2 A)\ 

If the same function of two angles are equal, the angles 
themselves must be equal, supposing them both in the 
same quadrant. 



90 - A = 2 A, whence A = 30 



27. tan B = cot (45 - § B) . 

28. sin 3 x = cos (2 x - 270). 

29. sec (2 x — 30 ) = esc (180 — x). 

30. cos — = sin (60——). 

4 \ 4/ 



Identities. 

Art. 11. It will be remembered that an identity differs 
from an equation in that the two terms are equal for all 
values of the unknown quantity, hence the two terms are 
exactly the same in value, but differ merely in form. In 
other words, the same relation is expressed in two different 
ways. For example, the words " sweat " and " perspira- 
tion " are two different ways of expressing the same idea. 

Likewise, tan x = . 

cos x 

It is often required for convenience or for simplicity to 

change the form of a trigonometric expression, and the 

fundamental relations already found, namely, sin 2 x+ cos 2 x 

sin x 

= 1, tan x = ; tan x . cot x = 1, sin x . esc x = i f 

cos x 

cos x . sec x = 1, make it readily possible to do this. 



The Relation Between Angle and Line. 169 

Example. Prove that esc 2 A (1 — sin 2 ^4) = cot 2 A. 

Since 1 — sin 2 A = cos 2 A and esc A = - — ! — 

sin A 

cos A 

.'. csc 2 A (1 — sin 2 A) = — = cot 2 A. 

V ; sin 2 A 

As a general rule, it is advisable to reduce all the terms 
of an identity to their simplest terms and to perform the 
indicated operations in order to show the equality. 

EXERCISE II. 
Identities. 

Prove the following identities : 

1. tan x sin x + cos % = sec x. 

2. sin A esc A tan A = tan A. 

3. sec 2 m esc 2 m sin 4 m + 1 = sec 2 m. 

4. 3 sin 2 x— 2 + 3 cot x cos x sin # = 1. 

_ 1 — 2 sin x cos x , x2 

6. ; = (sec x— esc xy. 

cos 2 # sin 2 x 



6. sin 4 x — cos 4 x + 2 cos 2 x — 1, 

1 + sin v. 



7 cos^ y 



sin ^ 

sin .4 

cos .4 tan 2 yl 



o sin ,4 . . 

8. = cot A . 



~ o , N sin 2 x 

9. cos J x (sec x + 1) h =- 2 cos #. 

sec 3; + 1 

iA cos n cot n — sin w tan n 

10. = 1 + sin « cos n 

esc w - sec n 



Trigonometric Equations. 
Art. 12. 

Example. Sec x + tan x = -n/J. (i) Find x. 
In order to solve this equation, since it involves the 
two unknowns, sec x and tan x, it is clearly necessary to 



1 70 Plane Trigonometry. 

have another independent equation involving the same 
unknowns. We can always find at least one equation 
between any two functions among the relations already 
established between the functions. In this case we have, 



tan 2 x + 1 = sec 2 x, 
or sec 2 x — tan 2 x= 1 (2) 




1 
Dividing (2) by (1); sec x - tan x = -—=■ 

V3 


(3) 


Adding (1) to (3) ; 




sec x + tan x = V3 




2 sec x = —=■ + V3 = ^V$ + V3 

V3 


= |V3 


.-. sec x =f \/J 
whence # = 30 . 





Another method which can be frequently used, is to 
express all the trigonometric quantities involved in the 
equation in terms of the same function of the unknown 
angle ; this method often results in a quadratic equation 
which can be solved in the usual manner. 

Suppose we have 2 V 1 — sin 2 A + sec A = 3 

then 2 cos A H = 3 

cos A 

2 cos 2 ,4 + 1 = 3 cos A 

2 cos 2 A — 3 cos A + 1 = o 

(2 cos A - 1) (cos A — 1) = o 

2 cos .4 — 1 = o also, cos A — 1 = o 

.*. cos /I = \ and cos ^4 = 1 

The angle whose cosine is \ is 6o°; the latter result 
where the cosine is 1 gives an angle o°. Hence we have 
two answers, namely, 6o° and o°. 



The Relation Between Angle and Line. 171 

EXERCISE III. 
Trigonometric Equations. 

Find the angle in the following equations : 

1. esc x = -f tan x. 

2. tan A + cot A = 2. 

3. sec 2 x + esc 2 x = 4. 

4. sin y + esc y = — f . 

5. 3 sin ^4 = 2 cos 2 ./l. 

6. sin 2 x = 2 - 



7. cot rr + sin rv 

8. sin 2 x + sin 2 x tan 2 # 




PART II. 



SOLUTION OF TRIANGLES. 



The Right-Angled Triangle. 

Art. 13. The process by which the unknown numeri- 
cal values of the parts of a triangle are computed from 
the known parts is called solving the triangle. 

By the use of the six trigonometric functions, any right 
r triangle may be completely solved when 
two of its parts, one of which is a side, 
are known. 

Given a right triangle MNR, right- 
angled at N. Also AM = 35 36' 
20" and n = 21.674 feet. Required AR 
and the sides r and m. 




54° 23' 4 o ; 



Fig. 8. 

I. To find R. 
v ZM + ZR = 9o°. .-. ZR = 90°- AM- 

II. To find r, having given M and n. 

Here it is necessary to use a formula which includes 
M, n, and r ; since n is the hypotenuse and r the side 
adjacent to M, the cosine of M is suggested. 

Now cos M = — , whence r = cos M X n. 



.'. log r = log n + log cos M = log 21.674 + log cos 
35° 35' 20" 

log 21.674 = 1-33594 
log cos 35 36' 20" = 9.91011 — 10 
log r = 11.24605 — 10 
= 1.24605 
r = 17.622 feet. 
172 



Solution of Triangles. 173 

III. To find m. 

It is clear that M, n, and m must occur in the formula 
selected; since n is the hypotenuse and m is the side 
opposite M, the sine of M is suggested. 

sin M = —, whence m = n sin M. 
n 

.'. log m = log n + log sin M = log 21.674 + log sin 
35° 36 r 20" 

log 21.674 = 1.33594 
log sin 35 36' 20" = 9.76507 - 10 
Add, log m = 1 . 1 o 1 o 1 

.*. m = 12. Gig feet. 

Given a right triangle ABC, right-angled at B. Also 
Z_A = 63 i2 r 25" and a = 11 2.34 feet. Required ZC 
and the sides b and c. 

I. To find ZC 

ZC = 90 - (63° 12' 25") - 26 47' 35". 

II. To find b. 

It is necessary to choose a function 
containing the given parts, A and a, and 
the required part, b. Since 5 is the hypot- 
enuse and a is the side opposite to A, the 
sine is suggested ; hence, A 

Fig. 9. 

sin A = — , whence b = and 

b sin /l 

log b = log a + colog sin A = log 112.34 + colog. sin 

6 3 ° 12' 25" 

log 112.34 = 2.05053 

colog sin 63 — 12' — 25" = 0.04932 

log b = 2.09985 

b = 125.85 feet. 




174 



Plane Trigonometry '. 



III. To find c* 

Here A and a are concerned, which suggests the tangent 

or cotangent. We may use either tan A = —ox. cot A = — . 

c a 

Since c is required, the latter is preferable ; 
whence, c = a cot A. 

log c = log a + log cot A = log 112.34 + log cot 
63° 12' 25", 



log 112.34 
log cot 63 12' 25" 



2.05053 

9.70328 — 10 



l ogc= 1. 75381 
log c = 56.730 feet. 



The Isosceles Triangle. 



Art. 14. Since a perpendicular from the vertex of an 
isosceles triangle bisects the base, the solution of an isos- 
celes triangle easily resolves itself into 
that of a right triangle. Let MNP be 
an isosceles triangle in which MN = 
PN. Drop the perpendicular NR ; then 
in the right triangle MNR, ZMNR 
= \ ZN, and MR = $ MP; hence if 
any two of the unequal parts, one of which 
is a side, be given, the unknown parts may be found. 

For example, suppose we have MN = x = NP = 
62.231" and ZN = 102 34' 12", to find MP = n and 
ZM - ZP. 

Since ZMNR = \ ZM, ZMNR= 51 17' 06". 

I. To find Z M. 

if = 90 - (51 - 17' - 06") = 3 d>° - 42 , -54 / '. 




Solution of Triangles. 



175 



II. 

MR 

hence 



To find MR. 
=\MP =\n 

sin — N = 
2 



here x, ^ N and ^ n are involved, 



1 • A^ 
or — n = x sin — ■ 

2 2 



•'• l°g (J n ) = ]°g x + l°g sm i ^ 
log 62.231 = 1. 79401 
log sin 51 17' 06 = 9.89224 — 10 
log (^ «) = 1.68625 

\n = 48-557 
w = 97.114. 



Solution of Regular Polygons * 

Art. 15. Since a regular polygon may be divided into 
isosceles triangles by lines from its center to its several 
vertices, its solution depends directly upon that of the 
isosceles triangle, and, therefore, ultimately upon the 
solution of a right triangle. For example : 

Let ABCDE be a regular pen- 
tagon ; from its center O draw 
OA, OB, OC, OD, and OE, to 
the vertices, thus dividing it 
into five isosceles triangles. A 
solution of one of these triangles 
will lead to a solution of all, 
and hence to a solution of the 
polygon. 

By geometry the sum of the angles at the center of the 
pentagon = 360 , and since the five angles there formed 
are all equal, each one, say AOB, is equal to £ of 360 = 
72 . In general, if the number of sides of a polygon be 

n, then each central angle will be ^ , or one half of each 




central anjrle will be 



180 



1 76 Plane Trigonometry. 

Drop a perpendicular OF from O to AB, then 

ZAOF=^- = ^- = 36°. 
n 5 

In general, calling the apothem of a polygon h, and 
one side of the polygon as AB, x\ then if the number of 
sides is represented by n, and the radius OA by r, we have 

. 180 k , i- # 

sin = 2 — whence r = — 

n r • 180 
sin 

n 

A I • l8o ° 

ana — x = r sin '■> 



Again, tan ^^ = i-^-, whence /j = — i-^ — , 
* tan ^ 

, 1 ^180° , 1 , , 180 

.°. h = — x cot and — x = h tan . 

2 n 2 « 

In the pentagon ABODE let ,45 = 9.7232 inches = #. 

Then Z.I AOB = ZAOF = — = 36 . 

5 
I. To find r. 

Here 3 6°, \ x and r, are involved in the right triangle 
AOF. Since r is the hypotenuse and \ x the side oppo- 
site the angle 36 , we have, 

sin 36 = i-^ , whence r = -A^- . 
r sin 36 

.*. log r = log % x + colog sin 36 = log 4.8616 

+ colog sin 36 

log 4.8616 = 0.68678 

colog sin 36 = 0.23078 



logr = 0.91756 
.*. r = 8.271 inches. 



Solution of Triangles. 

To find the perimeter : 

We had sin 3 6° = i^ 
r 

.'. % x = sin 36 r 

log sin 36 = 9.76922 

logr= -9i75 6 

log ^ x = 0.68678 

log 10 = 1. 00000 

1.68678 

.*. perimeter = 48.618 inches. 
Since the perimeter = 5 x?— 10 (^ x) = 
•'• l°g /> = l°g IO + l°g i x - 



177 



(suppose) 



To find /j : 

36 , i x and /z. are involved ; now tan 36 
h 



h 



or 



cot 36 



, whence h = \ x cot 36 ; .-. log h = log Ix 



+ log cot 36 


= log 4.8616 + log cot 36 




log 4.8616 = 0.68678 
log cot 36 = 0.13874 




log/* = .82552 
// = 6.6915. 




Areas. 


Art. 16. 


To find the area of a 



triangle. 

In the triangle ABC, call the base 
b, and the altitude // ; then by geom- 
etry, 

area ABC = £•& X h. (1) 

This formula applies when base and altitude are given. 

Again, in the right triangle ABD, sin /_ A = L (small 




i/S Plane Trigonometry, 

letters represent the sides opposite the angles indicated 
by large letters), whence h = c sin A. Substituting this 
value of h in (i), we get 

area ABC — \ be sin A. {2d) 

This formula applies when two sides and the included 
angle are given. It is evident, by drawing perpendiculars 
successively from the other vertices A and C, we obtain, 

area ABC = \ ac sin B (2b) 
and area ABC = \ ab sin C. (2c) 

A D TiC 

Again, in ABD, cot A = — , and in BDC, cot C = — 

h h 

.-. cot A + cot C = = — 

& h 



whence h = Substituting this value of 

cot A + cot C 

h in (1), we get, 

area ABC = ^ . (3a) 

cot A + cot C 



By drawing the altitude from A and C successively, it 
easily follows that, 

area ABC = * ^ (3b) 

cot 5 + cot C 

and area ABC = i^ . (3c) 

cot ^ + cot B 



This formula applies when two angles and the included 
side are given. 



Solution of Triangles. l 79 

The above formulae (3 a) , (3 b) and (3 c) will be found 
later to reduce to the forms : 

b 2 sin ^4 sin C a 2 sin i? sin C 
area ABC 



2 sin (.4 + C) 2 sin (5 + C) 
c 2 sin A sin B 
2 sin (4 + B) 

Example. Find the area of the triangle ABC, 
/_B = 32° 16' ; zlC = 25 39' and a = 23.27 inches. 

Here area ABC = - 



cot 2? + cot C 

h ( 2 3- 2 7) 2 



COt (3 2° 


16') 


+ 


COt (2 5 C 


'39') 


cot 32° 


i6 r 


= 


1-5839 




cot 25° 


39 r 


= 


2.0825 





cot B + cut C = 3.6664 

log area ABC = colog 2 + 2 log 23.27 + colog 3.6664 

colog 2 = 9.69897 — 10 
2 log 23.27 = 2.73360 
colog 3.6664 = Q-4357 6 ~ IO 
log area = 21.86833 — 2 ° 
- 1.86833 
area — 73.847 sq. in. 

It is evident, provided we have the necessary known 
parts, that the area of an isosceles triangle may always be 
found from formula 1, since the altitude and base may be 
calculated, if not given (see Art. 16). 

Area of the Regular Polygon. 

Art. 17. Since the regular polygon may always be 
divided into as many isosceles triangles as it has sides, 
by drawing radii to its vertices, its area is readily found 



i8o 



Plane Trigonometry. 



from previous formulae. Calling the perimeter, p ; one 
side, c ; and its apothem, h ; then by geometry, area of 
polygon = \ ph, and p = nc, where n is the number of 
sides ; also the angle of each isosceles triangle at the 

center equals ^— — or the half angle = 1 —^- . 

n 

With these data p and h can 
be calculated from the given 
parts. 

Example. To find the area 
of a regular hexagon, ABCFGH, 
given p = 74.116". Draw the 
perpendicular DE in the triangle 
BDC {D being the center). 




Z.BDE - 



i8o c 



3<^ 



74.116 



= 12.353. 



In the triangle BDE, cot BDE 



DE 
BE 



.-. DE = BE cot 



BDE. Now DE = h and BE = I c. .'. h = \ c cot BDE 



= \ c cot 30 . log h = log ^ c + log cot 30 
log cot 30 . 



log6.i76 + 



log 6.176 
log cot 30 



0.79071 
0.23856 



logh 
h 



1.02927 
10.697 



Area = ^ hp, whence log area = colog 2 4- log 10.697 
+ log 74.116, 



Colog 2 

log 10.697 
log 74.116 



9.69897 

1.02927 
1.86992 



10 



log area 
area 



2.59816 
396.43 sq. in. 



Solution of Triangles. 1S1 

It is also possible to find the parts of a regular polygon 
having a given area, since the central angles can always 
be obtained if the number of sides is known. 

Example. Find the perimeter of a regular decagon 
whose area is 336.72 sq. ft. 

Since area = \ hp and p = 10 c 

h hp = 33 6 -7 2 or 5 he = 336.72 (1) 

i c 
Calling the central angles each C, then tan 1 C= - — 

h 

1 c 

.'. h = — • Substituting this value of h in (1), 

tan \C 

1 c 

area = A = 5 c — % . .*. # c 2 = 336.72 tan i- C, hence 

tan \C 



■■ 134.688 tan 1 C= 134.688 tan 18 since 1C = ~^- 
8° ); whence 2 log c = log 134.688 + log tan 18 . 



log 134.688 = 2.12933 
log tan 18 = 9.51178 - 10 
2 log c = 1. 641 1 1 
log c = .82056 
c = 6.6154 



To find h. 



h = \ c cot I C = h (6.6154) cot 18 
log h = colog 2 + log 6.6154 + log cot 1 8° 

colog 2 = 9.69897 - 10 
log 6.6154 = 0.82056 
log cot 1 8° = 0.48822 

log // = 1 1.00775 — 10 
log h = 1.00775 
h = 10.18 



1 82 Plane Trigonometry. 

EXERCISE IV. 
Right Triangle. 
Solve the right triangle (right-angled at C), given : 

1. a = 2.3756, b = 6.1023. 

2. A = 29 13' 23", b = 27.132. 

3. B = 57 19' 32", c = 112.67. 

4. fr = .02567, a = .06211. 

5. a = 3-6378, A = 69 23' 45". 

6. The shadow of a steeple 102 ft. high is 116 ft. long. 
Find the elevation of the sun above the horizon. 

7. The guy-ropes of a derrick are 76 ft. long, and 
make an angle of 43 ° 25' with the ground. What is the 
height of the derrick, and how far from its foot are the 
guy-ropes anchored ? 

8. The elevation of a tower is 18 12' 16" at a dis- 
tance of 500 ft. What is the height? 

9. From a point A, immediately opposite a stake B, 
on the opposite bank of a river, a distance of 83.25 yards 
is measured to C at right angles to AB, and the angle 
ACB is found to be 62 ° 19/ 8". What is the breadth 
of the river ? ■ 

10. From the top of a lighthouse 98 ft. high, how far 
is it to the most remote visible point at sea, regarding the 
earth as a sphere 7918 miles in diameter? 

11. What is the angle of an inclined plane which rises 
1 ft. in 55 ft, measured horizontally? 

12. What must be the slope of a roof for a garret 
42 ft. wide, that the ridge may be 16 ft. above the garret 
floor? 

13. In a circle of 6.275 in. radius, what angle at the 
center will be subtended by a chord 10 inches long? 






Solution of Triangles. 183 

14. The angle between two lines is 44 32' 10". At 
what distance from the point of intersection will lie the 
center of a circle of 6235.2 ft. in diameter, tangent to 
both lines ? 

15. The diameters of two wheels, one on a shaft, the 
other on a machine, are 28 inches and 21 inches respec- 
tively, and their centers are 20 ft. apart. What length of 
belt is necessary for them ? 

16. Find the perimeter of an equilateral triangle cir- 
cumscribed about a circle whose radius is 13 inches. 

17. In the 15th example, what change would be neces- 
sary in gear wheel and belt to double the speed of the 
machine ? 

18. What is the length of i° on the circle of latitude 
through Pittsburg, latitude 40 27', if the radius of the 
earth (regarded as a sphere) is 3959 miles ? 

19. From the top of a hill the angles of depression of 
two stakes, set in straight line with the hill, 1200 yards 
apart, are observed to be 18 and 8° respectively. What 
is the height of the hill ? 

20. Two roads are non-parallel. From a certain point 
on one of them the angles between the first road and lines 
joining the point with two successive milestones on the 
other, are respectively 6° 30' and 12 15'. What is the 
distance between the roads ? 

21. Calling S the area of a right triangle, 

A = 31 20' 27", c - 211.89. Find 5. 

22. c = 12. 117, a = 9.208. Find S. 

23. S = 134.263, B = 33 12'. Find other parts of 
triangle. 

24. S = 32.73, c = 35.86. Find other parts of tri- 
angle. 

25. a = v 7 5, b = V$. Find other parts of triangle. 



Plane Trigonometry. 



EXERCISE V. 

Regular Polygons. 

Call the perimeter,^; apothem, h ; side, c; radius, r; 
Solve completely the regular poly- 

c = 2.7284. 

h = 9.2706. 

h = 23.819. Find n. 

h — 8.562. Find p. 

r = 6.768. Find h and p. 
a hexagon and an octagon are both 
302.64 sq. ft. Find the difference between their peri- 
meters. 

7. How many hexagonal tiles, 3 in. on a side, will it 
take to pave a hallway containing 225 sq. ft. ? 

8. The corners of a board 2 ft. square are cut away, 
leaving a regular octagon. What is the area of the octagon? 

9. A pentagonal fort is to have a diagonal of 500 ft- 
What will be the length of wall necessary to inclose it ? 

10. How many cu. ft. in the walls of a chimney in the 
form of an octagonal prism, if the apothem of a section is 
1', the thickness of the wall 4", and the height 40' ? 



number of 


sides, n . 


gons 


following : 


1. 


n = 


8, 


2. 


n = 


11, 


3. 


c = 


16.208, 


4. 


S = 


224.92, 


6. 


5 = 


196.22, 


6. 


The 


areas of 



PART III. 

FURTHER RELATIONS BETWEEN ANGLE AND LINE. 

Article 18. In Part I we have already discussed cer- 
tain relations between the trigonometric functions. We 
will now extend our investigations in this direction. It is 
customary to give the name goniometry to this branch of 
trigonometric analysis. 

In Trigonometry the same rules govern the direction of 
lines, which are already familiar to the student, through 
the graphical representation of equations in Algebra. 
There are, however, some further conventions which we 
will now explain. 

Definition : A directed line is one having a definite direc- 
tion by which it is distinguished. 

A line AB is understood as directed, and therefore 
measured from A towards B, while by BA we mean a line 
taken in the contrary direction, hence AB = — BA. 

We refer all points in the plane to two lines at right 
angles : the first, CA , is horizontal and is called the 
Abscissa or X-axis ; the second, DB, is vertical and called 
the ordinate or F-axis ; these cut at a point O, known as 
the Origin. 

Lines measured in the direction OA are positive. 
" " " " OB " positive. 

" " " " OC " negative. 

" " « " OD " negative. 



An angle is conceived as generated by a line revolving 
from its initial position OA, which coincides with the X- 

i35 



i86 



Plane IVigonometry. 



axis (Fig. 14), and extends along it towards the right. 
The revolution may take place either counter-clockwise or 
clockwise : in the former case the angle (AOA', Fig.' 14) 
is said to be positive, in the latter (AOA", Fig. 14), nega- 



<X 



/ 



B' 



B 

h+y 



■y 

D 
Fig. 14. 



.A' 



+ JC 

A" 



tive. The revolving line when fixed, bounds or terminates 
the angle, and therefore is often alluded to as the terminal 
line. 

The size of an angle is estimated from the horizontal 
diameter to the right of O, either counter-clockwise or 
clockwise to the terminal line, and can therefore be of 
any number of degrees up to plus or minus 360 , in one 
revolution, or may be made to contain as many positive 
or negative degrees as desired by repeated rotation of 
the revolving line. 

When a line such as OA' occupies any position between 
the X and Y axes we consider it as directed from O 
towards A', and that lengths measured from the origin O 
toward the extremity A ' are positive, while those taken in 
the opposite direction, as from O towards B', relatively to 
OA', are negative. This is true for all positions of a 
line between the X and Y axes, and hence we note in 
Fig. 14 that while OA" is positive, OC, being in the oppo- 
site relative direction, is negative. 



Flirt Jier Relations Between Angle and Line. 187 

Heretofore only the functions of angles less than 90 
have been considered. The question now arises whether 
the idea of the trigonometric functions can receive gen- 
eral extension. 

We have seen that by means of a revolving line an 



o/ 1 



y 




Av 


C 




X 


^ 


/ 


\ 






X 

B 










I 


I 





Fig. 15. 








Fig. 16. 



-X 



angle of any size can be obtained. Let us consider the 
functions of the angle XOA in each of the figures 15, 16, 
17, and 18. In every case the angle XOA is traced by 





Fig. 17. 



Fig. 18. 



the revolving line from the initial position, as shown by 
the curved arrow, to a terminal position 0/1. We notice 
that for angles greater than 90 we no longer, as hereto- 
fore, have a side opposite ; instead, however, we have a 
perpendicular drawn from the X-axis to the terminal tine, 
so that in each case considered we have a right-angled 
triangle formed by three directed lines BC, OB, and OC. 
Taking careful note of the directions of these lines, we 
obtain the following: 



TanXO^ = 



between 
Sin XOA 



1 88 Plane Trigonometry. 

Fig. 15. Fig. 16. 

A b g etfe^ °o_ 9 o* 90-I800 

Sin XOA - ^i+1 = + BC BC{ + ) _ BC 

OC ( + ) OC OC ( + ) oc 

CosXOA - 25J±i _ + QB OB{~) _ _OB 

OC ( + ) OC OC ( + ) OC 

BC ( + ) £C £C ( + ) _ BC 

OB ( + ) " OB OB (-) 0£ 

Fig. 17. Fig. 18. 

An v£l£^ 180° - 27 o° 27 o>- 3 6o° 

£C (-) __ __ BC BC (-) .. _ BC 

OC ( + ) " OC OC ( + )' ~ OC 

CosXOA - °*(-> - - ™ °^( + ) - + ^ 

OC ( + ) OC OC ( + ) OC 

TanlOi = * C <-> = + ^ y (-> - - ™ 
0£ (-) 0£ 0£ ( + ) OB 

Since the reciprocal of a trigonometric function has the 
same sign as the function, we can easily obtain the 
cotangents, secants, and cosecants, with their proper signs 
prefixed, from an inspection of the above table. 

It should be noted that to each positive angle XOA 
corresponds one negative angle of size 360 — X0^4,the 
trigonometric functions of which are exactly the same as 
those of the positive angle. 

Art. 19. Now as the function values do not depend 
upon the unit chosen, we might select some convenient 
length OC, on the revolving line, as our unit. 

We thus get, sin XOA = ^ = ^ = 30 (Fig. 15). 

OC 1 

Again, cos XOA = || = °^ = OB (Fig. 15). 

In the remaining positions of OC we will always 

have ± BC and ± OB representing the sines and cosines. 

This suggests that we should draw a circle of unit 



Further Relations Between Angle and Line. 189 



radius, with O as center, and endeavor, by suitable 
geometrical construction, to express the remaining func- 
tions as lengths, and not, as previously, by a ratio ; such 

E 




Fig. 19. 

a procedure would tend toward simplicity. Extending 
this conception, we will define the functions as certain 
lines determined by the angle involved in a circle of unit 
radius, known as a unit circle. 

Let DNMG be such a circle. With O as center, draw a 
horizontal and a vertical diameter. These two diameters 
divide the circle into four parts called quadrants, numbered 
I, II, III, IV. 

Suppose the radius stops in the position OC, then, since 
OD was the initial position, and OC is now the terminal 
line, the angle described is DOC. From C drop a per- 
pendicular upon the horizontal diameter OD, call it BC, 
and consider it as being directed from B towards C, or in a 
positive direction. 

At D draw a tangent DE ; let it cut the unit radius OC 
produced in H. Also draw a tangent NK at N, and let 
OC produced cut it at F. Then in the triangle OBC, 
representing /_ BOC by x, 

BC ( + ) 
OC (+) 

<m±l = + ob =+or 

OC ( + ) 1 



sin x 



+ ?£ = + BC 



cos X 



1 90 Plane Trigonometry. 

Again, in the right triangle OHD, 

0D{ + ) 1 

0H( + ) , OH ■ n „ 

sec x = — '- = -\ = + OH. 

OD (+) 1 

Now the two remaining functions, the cotangent and 
cosecant, will both be positive, since they are the recipro- 
cals of the tangent and sine respectively. In the triangle 
NOF, which is right-angled at N, we have Z.NFO = Ax, 

hence, cot x = ^- = ^- = NF 

NO 1 

and cosec x = -^L = -^^ = FO. 
NO 1 

We notice that all the signs of the functions in the first 
quadrant are positive, and further, that the three directed 
lines BC, OB, DH, give us by their directions the signs 
of the sine, cosine, and tangent, from which those of the 
remaining three functions, being the reciprocals of these, 
can be determined. 

If the angle is in the second, third, or fourth quadrant, 
a similar construction to the above will in each case 
enable us to express a function as a line of definite length 
and direction. Before, however, we endeavor to obtain 
these results, it will be necessary to define in general 
terms, and for any angle, the trigonometric functions 
referred to a unit circle. 



Definitions. 

Art. 20. The sine of an angle in a unit circle is the 
perpendicular to the horizontal diameter extending from 
it to the extremity of the moving radius. 

The cosine of an angle in a unit circle is the distance 
from the center of the circle to the foot of the sine along 
the horizontal diameter. 



Further Relations Between Angle and Line. 191 

The tangent of an angle in a unit circle is that part of 
the tangent to the circle at the right-hand extremity of the 
horizontal diameter, between the point of tangency and 
the point where the tangent intersects the moving radius 
produced forward or backward to meet it. 

The cotangent of an angle in a unit circle is that part 
of the tangent drawn to the circle at the upper end of the 
vertical diameter included between the point of tangency 
and the point where the tangent meets the moving radius 
produced forward or backward to meet it. 

The secant of an angle in a unit circle is the distance 
measured along the moving radius from the center to its 
intersection with the tangent. 

The cosecant of an angle in a unit circle is the distance 
measured along the moving radius from the center to its 
intersection with the cotangent. 

Art. 21. Keeping the above definitions carefully in 
view, the student will have no difficulty in obtaining tlie 
lines representing the functions of an angle in any 
quadrant, together with their signs, as given in the table 
below. (See Fig. 20.) 

Quad. I. Quad. II. 

sin DOC = + BC sin DOC - + B'C 

cos DOC = + OB cos DOC = - OB' 

tan DOC = + DII tan DOC = - DH' 

cot DOC = + NF cot DOC = - NF' 

sec DOC = + OH sec DOC = - OH' 

cosec DOC = + OF cosec DOC = + OF' 

Quad. III. Quad. IV. 

sin DOC = - B"C" sin DOC" = - B'"C 



tn r^m 



cos DOC = - OB" cos DOC" - + OB' n 

tan DOC = + DH tan DOC" - - DH' 

cot DOC" = + NF cot DOC" = - NF' 

sec DOC" = - Oil sec DOC" = + OH' 

cosec DOC" = - OF cosec DOC" = - OF' 



192 



Plane Trigonometry. 



Since the sine has the same sign in the quadrants that 
are side by side or on a horizontal line ; the cosine has 




the same sign in the quadrants lying on a vertical line, 
and the tangents have the same sign in the quadrants 
lying on diagonal lines, the result may be plotted thus : 





Fig. ax. 



Then if the sign of the cosine of any angle, for example, 
is in question, it is only necessary to observe whether the 
angle is in a quadrant on a vertical line with the first 
quadrant or not, etc. 



Further Relations Betweeti Angle and Line. 193 

It should be noted that these results are in accord as 
to sign with those in Art. 18 ; had we drawn our tangent 
lines at M or G this would not have been the case ; hence 
our reason for selecting D and N as points of tangency. 
The student should carefully note this fact. 

An analysis of the quadrants indicates that the varia- 
tions tabulated below take place among the functions 
while the angle is increasing from o° to 360 . 



Angle In- 
creases 
from 


— 90 


oo° — 180° 


180 — 270 


270 — 360 


Sine 


increases 
from to-f-i 


decreases 
from -(- 1 to 


decreases 
from to — 1 


increases 
from — 1 to 


Cosine 


decreases 
from -f- 1 to 


decreases 
from to — 1 


increases 
from — 1 to 


increases 
from to + 1 


Tangent 


increases 
from oto-f » 


increases 
from — 00 to 


increases 
from to + 00 


increases 
from — 00 to 


Cotangent 


decreases 

from -J- 00 to 


decreases 
from to — 00 


decreases 

from -foo too 


decreases 
from to — 00 


Secant 


increases 
from -j- 1 to-j- 00 


increases 
from — x to — 1 


decreases 
from — 1 to — 30 


decreases 
from + 00 to + 1 


Cosecant 


decreases 
from -j- 00 to -\- 1 


increases 
from+ i to+x> 


increases 
from — 00 to — 1 


decreases 
from — 1 to — 00 



To Express the Functions of any Angle in Terms of the 
Functions of an Angle in the First Quadrant. 

Art. 22. The values of the trigonometric functions 
are compiled in tables, which tables will be found to con- 
tain only angles in the first quadrant, or those between o° 
and 90 ; the reason for this lies in the fact that it has 
been found easy to reduce the functions of an angle in any 



194 



Plane Trigonometry. 



quadrant, to those of an acute angle ; this is of much 
practical importance, and will now claim our attention. 

To express the function of any angle in terms of the 
functions of an angle in the first quadrant. 

Let AOC = x be an angle in the first quadrant (Fig. 




Fig. 22. 



22), and CO A' an angle in the second quadrant, such that 
its supplement A'OC = x, then 180 — x = CO A'. 

Assuming a unit circle, and making the construction 
shown in Fig- 22, then, 



sin x = DA 
cos x = OD 
tan x = CE 



cot x = FG 

sec x = OE 

cosec x = OG 



Noting that /.CO A' = 

sin (180 - x) = D'A' 
cos (180 - x) = OD' 
tan (180 - x) = CE' 



180 — x, we have, 



cot (180 

sec (180 

cosec (180 



x) = FG' 
x) = OE' 
x) = OG' 



Now in the right triangles DO A and D'OA' we have 
OA = OA' (being radii), and the angle DO A is equal to 
the angle D'OA' (by construction), therefore these tri- 
angles are equal ; and taking notice of their directed lines, 



Further Relations Between Angle and Line. 195 

we see that D'A' = DA, and OD' = - OD (since their 
directions are opposite). Again, in the equal triangles 
COE and COE' we have CE' = - CE, and OE' = OE ; 
lastly, in the triangles FOG and FOG' which are also equal, 
FG f = - FG, and OG' = OG. 

Now by reference to the above values of the functions, 
we obtain : 

sin (180 — x) = + sin x cot (180 — .x) = - cot x 

cos (180 — x) = — cos x sec (180 — x) = — sec x 

tan (180 — x) = — tan x cosec (180 — #) = + cosec x 

The functions of (360 — x) can be easily obtained from 
Fig. 22, by drawing DN and considering ON as the ter- 
minal line of a reflex angle * CON in the fourth quadrant. 
The angle CON = x and the reflex angle CON = 360 - x. 
We thus have, sin (360-— x) = DN = — DA, or sin 
(360 — x) = — sin x, and in like manner for the other 
functions, see Art. 23. The remaining case where the 
angle is 180 + x can be obtained in a similar manner to 
the above, and is left to the ingenuity of the student. The 
results, however, are given in Art. 23. 

Art. 23. We will next consider an angle of (90 + x) 
in the second quadrant. Let COB = x be an angle in 
the first quadrant, and the angle COF' = 90 + x, the 
angle GOF r being equal to x. 

Making the construction shown in Fig. 23, we have, 



sin x = DB 


cot x = GF 


cos x = OD 


sec x = OE 


tan x = CE 


cosec x = OF 



Noting that Z COF' = 90 + x, we get, 

sin (90 + x) = D'B' cot (90 + x) = GF' 

cos (90 + x) = OD' sec (90 + x) = OE' 

tan (90 + x) = CE' ■ cosec (90 + x) = OF' 

* A reflex angle is one greater than 180 and less than 360 , and 
therefore in the III or IV quadrant. 



196 



Plane Trigonometry. 



In the right triangles DOB and D'OB', we have, 
OB = OB' and ZD'B'0 = ZDOB, .'. ADOB = AD'OB f . 
Further, D'B' = OD, and OD' = - DB. Again, in the 



F' G- 




Fig. 33. 



equal triangles EOC and F'OG we have OF' = 0£, and 
GF' = - CE ; while in the triangles GOF and COE', 
which are also equal, OE' = - OF, and CE' = - GF. 
Referring to the above values of the functions, we obtain, 



sin (90 + x) = cos x 
cos (90 + x) = - sin x 
tan (90 + x) = - cot x 



cot (90 + x) = — tan x 
sec (90 + x) = — cosec # 
cosec (90 + x) = sec a; 



By drawing KH and regarding the reflex angle COH 
as 270 + x, we may obtain in like manner the relations 
between the functions of (270 + x) and those of x. The 
case where the angle considered is (270 — x) involves a 
similar construction to the above, and should be investi- 
gated by the student. 

We will now present in tabulated form the results of 
the previous article. 



Further Relations Between Angle and Line. 197 



II Quad. 

sin (j8o°-x) = + sin x 

cos (180 — x) = — cos x 

tan (i8o°-x) = — tanx 

cot (180 — x) = — cot x 



III Quad. 

sin (i8o°-!-x)=— sin# 

cos (i8o° + x) = — cos# 

tan (i8o° + x) = + tan# 

cot (i8o° + x) = + cot# 

sec (i8o° + x) = — sec# 



sec (i8o°-x) = - sec x 
cosec (180 - x) = + cosec x cosec (180 + x) = — cosec x 



IV 


QUAD. 




II Quad. 






sin (360 - 


-*) = - 


sin a; 


sin (90 + y) = 


+ 


cosy 


cos (360 - 


- x) = + 


cos jc 


cos (90 + y) = 


- 


siny 


tan (360 - 


-*) = - 


tanx 


tan (90 + y) = 


- 


coty 


cot (360 - 


-x)=- 


cotjc 


cot (90 + y) = 


- 


tany 


sec (360 - 


- x) = + 


secx 


sec (90 + y) = 


— cosec y 


cosec (360 - 


-#) = — < 


:osec x 


cosec (90 + y) = 


+ 


secy 


II] 


! Quad. 




IV Quad. 






cos (270 - 


-30 - - 


siny 


COS (270 + y) = 


+ 


siny 


sin (270 - 


-30- " 


cosy 


sin (270 + y) - 


- 


cosy 


tan (270 - 


-y) = + 


coty 


tan (270 + y) = 


- 


coty 


cot (270 - 


-y)- + 


tany 


cot (270 + y) = 


- 


tany 


sec (270 - 


- y )= _, 


cosec 3; 


sec (270 + y) = 


+ cosec y 


cosec (270 - 


-3)=- 


secy 


cosec (270 + y) = 




secy 



These results may be epitomized in the following rules : 

1. Any function of an angle which is equal to 180 or 
360 plus or minus an acute angle, is equal to the same 
function of the acute angle, and will be positive or nega- 
tive according as the original function was positive or 
negative. 

For example, cos (180 — x) = — cos x. 

2. Any function of an angle which is equal to 90 or 
270 plus or minus an acute angle, is equal to the co- 



1 98 Plane Trigonometry. 

named function of the acute angle ; or, if the original 
function is a co-function, it will be equal to the plain 
function of the acute angle. The sign of the last 
function will agree with the sign of the original func- 
tion. 

For example, sin (270 — x) = — cos x. 

In both the above rules the sign of the final function 
is determined by the quadrant in which the original angle 
occurred. 

It will be noticed that any function of an angle greater 
than 90 can be reduced to the function of an angle less 
than 45 , or, if desired, to one of an angle between 45 ° 
and 90°, 

Art. 24. From the definition of negative angles it is 
evident that the moving radius, having described a nega- 
tive angle, will arrive at the same point as if it had de- 
scribed the positive angle, represented by 360 minus the 
number of degrees in the negative angle ; hence the 
functions of the negative angle and of this corresponding 
positive angle will be exactly the same. 

For example, the functions of - 75 are exactly the 
same as the functions of + (360 — 75 °) = + 285. 

Therefore, to find the functions of a negative angle, 
subtract this angle (as if it were positive) from 360 , and 
find the functions of the positive angle resulting. 

For example, tan — 125 = tan (360 — 125°)= tan 
235 = tan(2 7 o°- 35 ) = cot 35 . 

Art. 25. In Art. 18 we drew attention to the impor- 
tant part played by the directed lines of a right triangle 
in determining the trigonometric functions. Up to 
this point we have considered the initial line as horizon- 
tal, but this is often not the case, as a triangle may obvi- 
ously occupy any position in the plane. To facilitate the 
recognition of the functions of an angle in varying posi- 
tions, we suggest the following rules : 



Further Relations Between Angle and Line. 199 

I. The perpendicular is always at right angles to the 
initial line, or the initial line prolonged backwards ; it is 
directed from the initial to the terminal line, and is posi- 
tive for acute and obtuse angles, and negative for reflex 
angles. 

II. The hypotenuse is invariably directed from the 
vertex to the perpendicular, and is positive in all positions. 

III. The remaining side, the adjacent side, if the 
angle be acute, is directed from the vertex to the foot of the 
perpendicular. It is positive for acute angles and reflex 
angles between 270 and 360 , and negative for angles 
between 90 and 270 . 

In giving the ratio expressing any trigonometric function 
of an angle, care should be taken to give the correct 
directions of lines to which reference is made. 




In Fig. 24, keeping the above rules in mind, we have: 

cos EBK = — , cos FED - ^- , cos FGD - -^ 
BK ED GD 

D TT 

cos GDH = . Note that in each case the directed 

DG 
lines are measured from the vertex to the perpendicular, 

EK 
BE' 



according to Rules II and III. Again, tan EBK 



tan FED 



- FD -, tan FGD = -^-, tan GDH = S$- 
EF GF DH 



200 Plane Trigonometry. 

Further, in triangle DBM , tan DBM = ^- , while cot 

BD 

DKG in the triangle DKG = ^- . Taking the obtuse 

DG 

angles BKG and BKM, we have tan BKG = DG , and 
5 KD 

tan BKM = ^- . By Rule III the latter functions are 
KD J 

negative. 

Art. 26. In articles 22 and 23, we found certain 
relations existing between the functions of an angle formed 
by increasing or diminishing 90 , 180 , etc., by an acute 
angle x, and the functions of x ; we obtained, for example, 
sin (x + 90) = cos x. It would seem natural to inquire 
at this point, whether it would not be possible to increase 
or diminish x by any other angle y, and find relations 
between the functions of the new angle (x ± y) and those 
of x and y. A geometrical investigation of such a prob- 
lem clearly includes several cases, according as the ter- 
minal lines of x and y lie in the several quadrants ; it 
will be found convenient at first to confine our attention 
simultaneously to the two cases where (x + y) <9o and 
(x + y) > 90, in the latter of which the angle x is acute 
and y extends into the second quadrant. 



To express sin (x + y) and cos (x + y) in Terms and 
Functions of x and y. 

Art. 27. In each of the Figs. 25 and 26, let the 
angle DBC = x and the angle CBN = y ; in Fig. 25, 
the sum of these angles, x and y, is less than 90 , while 
in Fig. 26, it is greater than 90 , but the individual 
angles x and y are each less than a right angle. 

In both cases take a point A , upon the line bounding the 



Further Relations Between Angle and Line. 20 1 

angle y, and let fall two perpendiculars, one AG, upon 
the initial line of x (or the initial line produced backward, 




as in Fig. 26) ; the other AE, upon the terminal line of x. 
Note that the angle included between these perpendiculars 
to the sides of the angle x is GAE = x (by Geometry). 




Now in each case we have a right-angled triangle GBA 

formed by the line AG and sin DBA = sin (x + y) = ^-> 

BA 



202 Plane Trigonometry. 

Again, the perpendicular AE completes a right triangle 
AEB which contains the angle EBA = y. 

We have, however, as yet, no right triangle with x as an 
angle. Now at least one such triangle must be constructed, 
if it is our purpose to establish relations between the sides 
of the triangle ABG, whose ratios represent the functions 
of (x + y), and line ratios representing functions of the 
individual angles x and y. This suggests our drawing 
EF _L BD and EH J_ GA . We thus obtain two new tri- 
angles BFE and A HE, the former right-angled at F, the 
latter at H, each of which contains an angle x. 

It now remains to investigate geometrically the relations 
existing between the sides of the triangle ABG, whose 
ratios represent the functions of (x + y), and those of the 
triangles ABE, BFE, and A HE, giving careful attention 
to the directed lines. 

We have, in each figure, sin (x + y) 



BA 

n A nTJ i TT A T?T? i TJ A 

Now, 



GA 


GH + HA 


FE + HA 


FE HA 


BA 


BA 


BA 


= BA + BA 



Note that in these ratios each numerator and denominator 
is the side of a triangle containing an angle equal to x or 
y ; it was, in fact, to obtain this result that we broke up 
GA. Again, FE is a side of a right triangle FBE, and 
BA of a right triangle ABE ; both these have a common side 

BE: hence if BE is introduced into the ratio , by 

BA J 

breaking it up into two ratios without altering its value, 

thus, - — . — = — - . — ■ , the former will be a function 
BA BE BE BA 

of x, namely, sin x, and the latter a function of y, or, as we 
see, cos y. 



Further Relations Between Angle and Line. 203 

HA 
Again, taking the ratio , we note A H is a side of 

the triangle A HE, and BA of BAE ; these have a common 
side AE ; introducing this as above, and observing the 
directed lines, we get, 

AH AH EA AH EA 

= • — = . — - = cos x sin y. 

BA BA AE AE BA ' 

Collecting our results, we finally obtain, 

sin (x + y) = sin x cos y + cos x sin y. 

We have discussed this proof at length because of its 
great importance. The steps are few and simple, but it is 
desirable that the reasons for them should be clearly 
understood by the student before proceeding further, as 
other similar proofs follow which form a basis for many 
of the most important formulae in trigonometry. 

For the sake of clearness we will now give concisely 
the necessary steps and construction discussed above. 

Art. 28. I. To slww that sin (x + y) = sin x cos y 
4- cos x sin y. Let 

ZDBC = x and ZCBN = y ; then Z DBN = x + y. 

Take a point A on BN ', the bounding line of y, and draw 
AG±BD, AE±BC, also EH ± AG and EF _L BD. 
The angle GAE = x } v its sides are _L to those of x- 



sin (x + y) 



sin (x + y) = sin % cos y + cos x sin y. . . (1) 



GA _ 
BA ' 


GH+HA 
BA 


FE + HA 
BA 


FE HA 
BA BA 


FE 

BE ' 


BE + AH 
BA AE 


EA 

' BA' 





204 



Plaiie Trigonometry. 



II. Ttf show that cos (x + y) — C0s # C0J y — sin x sin y } 
making the same construction as above, we get, 



cos (x + y) = 



Fig. 27. 

BG 
BA 
BF 



GF 



BA 




( GF-BF " 




Fig. 28. 
cos (x + y) = — 



BG 



BA 

BF-GF 

BA 



Fig. 28, 



Hence in both cases, 






, , * BF-GF BF 

cos (x + y) = = — ■ - 

v n BA BA 


HE 
BA 


(Notice again the 
use of the com- 


BF BE HE 


EA 


mon side of the 


BE BA AE 


BA" 


triangles.) 


.*. cos (x + y) = cos x cos y 


— sin 


x sin y. . . (2) 



Further Relations Between Angle and Line. 205 

III. To show that sin (x — y) = sin x cos y — cos x sin y. 
Given Z x < 90. Let Z DBC = x and Z CBN = y, then 
DBN = x - y. 




Making the construction indicated in Fig. 29, and 
observing that the point A is again taken upon the line 
bounding the angle y ; also that the Z KEA = x, we have, 



• ( , GA KF KE - FE 

sin (x — y) = — = = 

y) BA BA 



BA 



KE 
BA 
KE 
BE 



FE 
BA 
BE 
BA 



EF 
EA 



EA 
BA 



\ sin (x — y) = sin x cos y — cos x sin y. 



(3) 



IV. To show that cos (x — y) = cos x cos y + sin x 
sin y. Referring to Fig. 29, we see that 

BG BK + KG 



cos (x - y) = 



.'. cos (x - y) 



BA 
BK 
BE 



BA 



BK FA 
BA BA 



BE + FA 
BA EA 



EA 

BA 

cos x cos y + sin x sin y. 



(4) 



206 



Plane Trigonometry. 



Art. 29. It might now occur to the student to inquire 
whether the proofs of the four formulae just given could 
not be simplified by making use of the unit circle, and, 
should we attempt such a method, which line would lend 
itself best to selection as a unit radius ? A glance at the 
four cases just investigated shows that the line extending 
from the vertex B to the point A, selected upon the line 
bounding the angle y, always appears as a denominator, 
hence our choice obviously lies with this line. Making 

if 





the same construction as before for case I, and in addi- 
tion describing an arc through A with B as center, further, 
putting BA = 1, we obtain 

sin (x + y) = GA = GH + HA = FE + HA. 
Now 



and 
but 



in x = 


BE 


.-. FE = 


sin x 


BE 


OS x = 


AH 
AE 


.-. AH = 


cos X 


AE 


BA - 


1 








BE = 


cos y 


and AE = 


sin y. 





Substituting these values, we get 

FE = sin x cos y and AH = cos x sin y. 
Hence sin (x + y) = sin x cos y + cos x sin y. 



Further Relations Between Angle and Line. 207 

A similar method may be adopted in each of the other 
cases. 

Art. 30. We might now ask, Are the formulae that we 
have just derived, general truths ? Do they hold when 
the dimensions of the angles x and y are unrestricted ? 
Also when either or both are negative ? 

The student will readily see that it would be tedious to 
attempt to investigate all possible cases geometrically. 
A more simple method is to show analytically that the 
formulae still hold good if the angles x and y be increased 
by a right angle, or if equal negative angles be substi- 
tuted, and hence establish the general truth. 

Thus, sin \ (x + 90) + y\ = sin J 90 + (x + y) \. 
By Art. 18, sin (90 4- A) = cos A. 

If A = (x 4- y) we get 

sin {90 4- (x 4- y) \ = cos (x 4- y) = cos x cos y 
— sin x sin y. 

Now cos x = sin (90 4- x) and sin x = — cos (90 4- x). 

Substituting, we get 

sin {(90 + x) 4- y} = sin (90 + x) cos y 
4- cos (90 + x) sin y. 

From the above equation it follows that if sin (x 4- y) 
= sin x cos y + cos x sin y be true for any special quad- 
rant, it also holds when x is in the following quadrant. 
But this equality has been proved for the first quadrant, 
hence it is true when x is in the second, and thus the 
limit may be indefinitely increased. A similar method of 
procedure shows that if either or both the angles be in- 
creased at will, or negative angles substituted, the truths 
expressed by the four formulae of the previous article re- 
main — and hence they are universal. 



208 Plane Trigonometry. 

Art. 31. From the relation tan A = , it is easily 

cos A 

possible to find an expression for tan (x + y) as follows : 

. „ , , x sin (x + y) sin x cos y + cos x sin y 

tan {x + y) = -. — = : : — - 

cos (x + y) cos x cos y — sin # sin y 

Divide by cos x cos y ; 

sin x cos y cos 5c sin y 

. / , x cos x cos y cos # cos y 

tan (x + y) = z : r- -^ 

cos x cos y _ sin x sin y 

cos x cos y cos x cos y 

, , N tan x + tan y , N 

.-. tan (# + y) = L . . ..... (5) 

" 1 -tan ^tany v/ 

Likewise, 

. , , N cos (x + y) cot x cot y - 1 „ N 

cot (x + y) = -^—j ^ = z ■ . . . (6) 

sin (x + y) cot y + cot x 



Tan (x — y) and cot (x — y) are found in identically 
the same manner as tan (x + y) and cot {x + y). 
Art. 32. Suppose in the formulae: 

sin (x + y) = sin x cos y + cos x sin y 
cos (x + y) = cos # cos y — sin x sin y 

, „ ( , N tan* + tany 

tan {x + y) = £- 

1 — tan x tan y 

j . , x cot x cot y — 1 

and cot (# + y) = z 

cot x + cot y 

y be made equal to x, then these formulae become, 
sin 2 * = sin x cos x + cos * sin x = 2 sin x cos x . (7) 



Further Relations Between Angle and Line. 209 

cos 2 x = cos 2 x — sin 2 x (8) 

, 2 tan x , s 

tan 2 # = (9) 

1 - tan 2 a; 

j , cot 2 X — I / n 

and cot 2 x = — (10) 

2 cot x 



\i 2 x = Am these formulae, then x — \ A , and substi- 
tuting we have, 

sin ^4 = 2 sin J ^4 cos \ A . . . (7) 

cos ^4 = cos 2 J ^4 - sin 2 \ A . . (8) 

, 2 tan A /I , N 

tan A = ; (9) 

1 - tan 2 i A W/ 

, . . COt 2 A yl - 1 , v 

and cot A = (10) 

2 cot \ A 

That is, the sine of any angle equals twice the product 
of sine and cosine of half the angle. The cosine of any 
angle equals the square of half its cosine minus the square 
of half its sine. Write the corresponding rules for tangent 
and cotangent. 



Functions of Hal} an Angle. 

Art. $5. In the formula cos A = cos 2 \ A — svi\ 2 \A 
since cos 2 A = 1 - sin 2 \ A 

cos A = 1 - sin 2 \ A - sin 2 J A = 1 — 2 sin 2 \ A 
whence 2 sin 2 \ A = 1 — cos ^4 

oi4 I COS ./I / v 

or, sin- \ A = (11) 



2 1 o Plane Trigonometry. 

Making the contrary substitution, sin 2 \ A =• i — cos 2 \ A ; 
cos A = cos 2 J A — (i — cos 2 J A) = cos 2 J ^4 — i + cos 2 
£ A = 2 cos 2 J ^4 — i, 

whence, 2 cos 2 J ^4 = 1 + cos A , 

? ^ A 1 + cos ^4 / v 

or, cos 2 J A = — — (12) 



Since tan J 4 = sm * A 

cos A ^4 



2 






sin i A 






cos J A 






1 — cos ^4 






2 


1 - 


- cos A 



tan 2 i 4 = s _H!i_£i = _ = 1 - ^» /1 . (13) 

cos 2 %A 1 + cos A 1 + cos ^4 

2 

and cot* J ^ = ' + cos A (14) 

1 — cos A 

These formulae make it readily possible to find the 
functions of half an angle when any function of the whole 
angle is given. 

For example, to find the functions of 30 having 
tan 6o° = V3 given. 

From the formulae, tan 2 6o° + 1 = sec 2 6o° 
it follows that, 3 + 1 = sec 2 6o° 

sec 6o° = 2 
hence, cos 6o° = J. 

From (11) and (12) are derived respectively, 

• o 1 — cos 6o° 1 — A , 

sin 2 30 = = 2- = \ 

2 2 

whence, sin 30 = \ 

o 1 + cos 6o° 1 4- 4- o 

cos 2 30 = = *■ = f 

2 2 

.-. cos 30 = -IV3 



Further Relations Between Angle and Line. 211 

, o 1 — cos 60 1 — \ \ , 

tan' 5 30 = = 1 = = i 

1 + cos 60 1 + \ f 

.*. tan 30 = Vj = \ V3, etc. 

It will be observed that these results tally with those 
obtained in Art. 9, page 10. 



Sum and Difference of Functions. 
Art. 34. Returning to the formulae, 

sin (x + y) = sin x cos y + cos x sin y . (1) 
sin (x — y) = sin x cos y — cos x sin y . (3) 

If (x + y) is replaced by a single angle, say P, and 
(# — y) is replaced by another angle, say Q, then the 
addition of the two formulae above will give the sum of 
the sines of P and Q, thus : 

x + y = P 

x - y = Q 

add, 2 x = P + Q 

subtract, 2 y = P - Q 

Adding (1) and (3) above, 

sin (x + y) + sin (x - y) = 2 sin # cos y. 

Substituting values of x and v, x + y, and # - y, assumed 
above,. 

sin P + sin Q = 2 sin i (P + Q) cos ,V (P - Q). (15) 
Subtracting (3) from (1), and substituting, 

sin P - sin Q = 2 cos £ (P + Q) sin J (P - Q). (16) 
In the same way, taking (2) and (4), 

cos (x + y) = cos x cos v — sin x sin v . (2) 
cos (x — y) = cos x cos y + sin x sin y . (4) 



212 Plane Trigonometry. 

Adding and substituting, 
cos P + cos Q = 2 cos J (P + Q) cos i (P - Q) . (17) 

Subtracting (4) from (2), and substituting, 
cos P - cos Q = - 2 sin 1 (P + Q) sin 1 (P - Q) (18) 

Exercise. State formulae (15), (16), (17), and (18) as 
rules. 

EXERCISE VI. 

Goniometry. 

Express the following functions in terms of the functions 
of angles less than 45 °. 

fsin 113 ; cos 216 ; tan 97 ; tan 315 ; cot 263 ; 

1. -|sec 190 ; esc 181 ; cos 302 ; sin 220 ; tan 175 ; 
[cot 316 ; cos 156 ; cot 142 . 

2. sin - 72 ; cos - 118 ; tan - 217 ; cot - 105 . 

By application of formulae find the simplest value of the 
following function : 

3. sin (180 — x) ; cos (180 + x) ; tan (90 + x). 

4. sin (270 + y) ; cos (270 - y) ; sin (360 - x) ; tan 

(360 - y). 

Prove following relations : 

- . / x 1 — tan x 

5. tan (45 - x) = . 

1 + tan x 

6. cos 3 x = 4 cos 3 x — 3 cos x. 

7. sin 3 x = 3 sin x — 4 sin 3 x 

8. sec A esc A = 2 esc 2 A. 

9. cot J y + tan \ y = 2 esc y. 

10. sin # = ¥ 9 T ; cos x = %\ ; sin y = -^ • cos y = ^f : 
Find sin (x + y) and cos (x — y). 



Further Relations Between Angle and Line. 213 

11. cos 26 = .9. Find sin 13 and cos 13 . 

12. sin 53 8' = .8. Find sine, cosine, and tangent 
of 106 16'. 

By formulas (15), (16), (17), and (18) show that, 

13. sin (45 + x) + sin (45 - x) = 2 sin 45 cos x 

= v 2 cos x. 

14. sin (150 + x) — sin (90 — x) = 2 cos 120 sin (30 

+ x) = - sin (30 + x). 

15. If x, y, and z are the three angles of a triangle, prove, 
sin x + sin y + sin z = 4 cos ^ x cos \ y cos ^ z. 

Transform the following into expressions suitable for 
use of logarithms : 

16. tan x + tan y. 19. cot x cot y — 1. 

17. cot x + tan x. 20. sin x — cos 2 x sin # 

sin x 

18. tan x tan y + 1. 21. 



V: 



COS X 



22 1 - cos 2 /I 23 tan x + tan y 

1 + cos 2 ^4 cot x + cot ;y 



24. 



1 — tan 2 x 
1 + tan 2 x 



Inverse Trigonometrical Functions. 

Art. 35. In an algebraic equation involving more than 
one unknown quantity the equation may be solved for any 
one of the unknowns in terms of the others, thus : if 3 x y 
= 4, solving for x, x — | y ; or for y, y = | x. 

Likewise in Trigonometry the expression, y =. sin % 
may be solved for x, by adopting a notation like this : 

x = sin -1 v, read x is the angle whose sine is y, or 
x = anti-sine of y. 



214 Plane Trigonometry. 

The symbol — i above the sine symbol must not be mis- 
taken for an exponent, although it is adopted from the 
analogy of the process to exponential division : sin -1 is 
a single symbol and inseparable, having a definite mean- 
ing, distinct from sin. 

Likewise, we have cos -1 A, tan -1 A, cot -1 ^, etc.; 
read respectively anti-cosine, anti-tangent, etc. 

Art. 36. These inverse functions may be readily con- 
verted into direct functions by setting them equal to 
another quantity, representing their value. 

For example, sin -1 J = 30 

whence -J- = sin 30 

or, in general, sin -1 A = B, whence A = sin B. 

It is to be observed that the expression sin -1 A rep- 
resents the angle and not its function, namely, the angle 
whose sine is A, which above is called B. 

Example. Prove cot -1 a + cot -1 b 



cot 



_! ab 



a + b 
Let cot -1 a = x, hence a = cot x 

and cot -1 b = y, hence b = cot y. 



Substituting in formulae cot (x + y) = - — - - 

cot x + cot y 

cot (cot- 1 a + cot" 1 b) 



or cot -1 a + cot -1 b = cot -1 



a + b 

ab — 1 
a + b 



PART IV. 

SOLUTION OF OBLIQUE TRIANGLES. 

Article 37. In the right triangle the right angle is 
always known, and it is always possible to find remain- 
ing parts, when two are given, provided the two given 
parts are not two angles. In the oblique triangle there 




Fig. 32. 



are in general six variable parts, three of which must be 
known, in order that the triangle may be completely 
solved. 

It is plainly necessary to divide the oblique triangle 
into two right triangles, that the relations of its parts 
may be found, through the medium of angle functions, 
since these latter are defined as ratios in right triangles. 
In the triangte ABC, then, draw the perpendicular BD, 
and call the sides a, b, and c, using small letters for the 
sides opposite the angle denoted by the corresponding 
large letters. 



215 



2 1 6 Plane Trigo?iometry. 

In the right triangle ABD, 



smA=- (where BD = h) {a) 



c 



In triangle BDC, sin C = - ^ 

Divide (a) by (b) $^ - * ...'.. (19) 

sin C c 



By drawing perpendiculars from the other vertices, 
successively, in the same manner may be shown, 

sin A a , s 
(20) 



sin B b 
sin B b 



sin C c 



(21) 



With (19), (20), and (21), if two angles and the in- 
cluded side are given, the remaining parts may be found. 

Art. 38. A slight transformation produces a formula 
which makes it possible to find the unknown parts when 
two sides and the included angle are given. 

Taking (19) by division and composition, according to 
the theory of proportion, 



sin A — sin C a — c 



sin A + sin C a + c 
Dividing (16) by (15), 

sin P - sin Q _ 2 cos j (P + Q) sin \ (P - Q) 
sin P + sin Q ~ 2 sin 1 (P + Q) cos \ (P - Q) 

= cos h (P + Q) sin j (P - Q) 
sin 1 (P + Q) cos 1 (P - Q) 

= coti(P + <2)tan| (P_0. 



(») 



Solution of Oblique Triangles. 217 

Replacing P and Q by A and C respectively in this 
formula, 

cot J, (A + C) = tan£ (A - C). 



sin A + sin C 



Substituting this value of — : in (m), 

sin A + sin C 

cot $ (A + C) tan \ (A - C) - -£-=^ 

a + c 

or tan £ (4 - Q = ^-^tan 1 (^ + C) . (22) 

a + c 

Since 4 + C = 180 - B, 



this formula makes it possible to find the remaining 
parts when the two sides a and c and their included 
angle B are given. 

By an exactly analogous process, using (20) and (21), 
may be derived, 

tan i (A - B)= ^|tan \ {A + B) . . (23) 

and tan * (5 - C) = -^-J— ^tan l (B + C) . . (24) 

+ c 



which meet all requirements, when any two sides and 
their included angle are given. 

Art. 39. In the above case the third side may be 
found directly, without finding the two unknown angles, 
by employing the geometrical theorem relative to the 
square of a side opposite an acute angle. 



218 



Plane Trig07iometry. 



If ABC is any triangle, BD being a perpendicular from 
2? upon AC, then by geometry, 

c 2 = a 2 + & 2 - 2 6 x £>C, 



but, in the right triangle BDC, cos C 



DC 



cos C. 

Substituting this value of DC above, 

c 2 = a 2 + b 2 — 2 ab cos C . 



or DC= a 



• (*5) 




Fig- 33- 

By drawing perpendiculars from the other vertices and 
applying the same theorem, are obtained the following : 

a 2 = b 2 + c 2 - 2 be cos A . . . . (26) 
b 2 = a 2 + c 2 - 2 ac cos B . . . . (27) 

Art. 40. There is plainly a third case that arises in 
the solution of oblique triangles, namely, when the three 
sides are given. 

a 2 + b 2 - c 2 



Solving (25) for cos C, cos C 



2 ab 



Subtracting each side from 1 and then adding each 
side to 1, are obtained, 



1 — cos C = 1 — 



a 2 + b 2 - c 2 2 ab - a 2 -V 



2 ab 



2 ab 
c 2 - {a - b) 2 
2 ab 



(K) 



Solution of Oblique Triangles. 219 

a 2 + b 2 - c 2 2 ab + a 2 + b 2 - c 2 



i + cos C = 1 + 

2 ab 2 ab 

(n _l_ A\2 _ a 

■ (S) 



2 ab 2 ab 

m (a + b) 2 - c 2 
2 ab 



Factoring (R) and (S) , observing that the numerator of 
the right-hand members of both equations are each the 
difference of two squares, 

1 - cos C = \c-(a-b)'\[c + a-b'] = (c-a + b) (c + a-b) 
2 ab 2 ab 

j + COS C - !« + *-') (° + b + C ) . 

2 ab 
From (11) and (12), 

1 - cos C - 2 sin' i C = JLZJL+D fe ± g ~ ^ W 

7 2 aft W 

1 + cosC - 2 cos' iC- (fl ^" f) (a + b + C ) ' W 
2 2 a& w 

Putting s = i (a + b + c) or 2s=a + b + c, 
then, 2(^-a)=<;-a + 6 

2 (5 — b) = c + a - b 
2 (s — c) = a + b — c 

Substituting these values in (u) and (v) , 

2 sin 2 £ C = 4 (•? - a) - ft) = 2 (s - a) (5 - b) 
2 ab ab 

2 COS 2 $ C = 4 * (5 ~ g ) - 2 ' ( * - C ) 

2 a£> a6 



or, sin iC-\/ (f -"><'-»> (28) 

i C = y/ f ( * ~ g ) . (29) 



COS 

ab 



220 Plane Trigonometry. 

Divide (28) by (29), 

taniC = v/^^?^ ■ 
V s (s - c) 



(30) 



By an exactly analogous process, corresponding expres- 
sions for ^ A and ^ B are found as follows : 



/ is -b){s- c) 
be 



sin \ A = 1 

= 4 his - a) 



cos \ A 



be 



tan 



t s (s — a 



(s -c)(s- b) 
) 






sin 



cos i B = 






*) 



tan i 5 



= t/ O - a) (^ - 



A comparison of (30), (33), and (36) will show that, 



(3i) 
(32) 
(33) 
(34) 
(35) 
(36) 



V 



(5 - a) (5 - b) (s - c) 



is a common factor. 



If this expression be represented by r, 
then (30) may be written, tan \ C 



and 



(^^) may be written, tan \ B = — — 

s — a 

(36) may be written, tan \ B 



s-b 



which reduces the calculation of the three angles to the 
determination of the value of one radical expression, r. 



Solution of Oblique Triangles. 221 

EXERCISE VII. 

Oblique Triangles. 

Calling the angles A, B, and C, the sides respectively 
opposite a, b, and c, and the area S, solve the following 
triangle : 

1. A = 6 9 °2i , 3o // , C=2 3 ° 11' 17", a = 123.23. 

2. 5 = 101 42' 21", ^ = 47° 12' 19", 5 = 10.029. 

3. B = 99 12' 10", C = 35 o' 40'', a = 1027.2. 

4. A - ii° 17' 33", 3 = 77°i5 / > ^ = 3-4576. 

5. ^4 = 82 12' 36", 6 = 62.117, c = 90.741. 

6. B = 109 49' $&", a = 22.222, c = iq.34. 

7. C - 67 58' 58", a = 393-6H, c = 208.47. 

8. B = 23°27 / 5o", 6 = .08679, a = -07241. 

9. a = in, £ = 425, <; = 238. 

10. a = 1023.75, b = 978.36, c - 1321.13. 

11. a = 18.705, b = 23.202, c = 9.667. 

12. Find S in each of the above examples : 

13. A line AB, 225 yds. long, is measured off on level 
ground. The angles formed with it by imaginary lines to 
C, a point in the same plane, are respectively 98 12' 23" 
and 78 9' 21". Find the distance from A to C. 

14. In running a line from B to C, two points in a 
survey, an impenetrable swamp is encountered. A third 
point D is chosen, from which B and C are both visible 
and accessible. The distances DB, DC, and the angle 
CDB are then measured and found to be, DB = 429.58 
ft., DC = 319.26 ft., and Z.CDB = 18 21' 36". Find 
length and direction of BC. 

16. Two forces of 1J6.5 and 200 pounds per sq. in. 
respectively make an angle of no° 25' with each other. 
Find the intensity and direction of their resultant. 

16. Three forces of 95.265, 68.21, and 105.2 lbs. 
respectively are in equilibrium. Find the angle between 
the first two. 



222 Plane Trigonometry. 

17. To find the height of a steeple, a line mn ioo ft. 
long is measured on the ground, and the horizontal angles 
at m and n made by mn with imaginary lines drawn to the 
point directly below the top of the steeple on the ground, 
are found to be 8o° 9' 25" and 72°3i r 13" respectively. 
Also the elevation of the top of the steeple from m is 
i4°2 / 3o // . What is the height of the steeple ? 

18. From a point in a 2 5 % slope, the angle subtended 
by a tower higher up the slope is 29 16' 25". From a 
point 75 feet higher up it subtends an angle of 42 ° 12' 
i*]". Find height of tower. 

To Express Angles in Radians. 

Art. 41. In addition to the unit of angular measure, 
the degree, used in Geometry, Trigonometry employs a 
unit called a radian. 

A radian is the central angle, in any circle, whose arc 
is equal in length to the radius. 

Hence the number of radians in a given angle is the 
number of times its arc contains the radius of the circle 
at whose center its vertex is placed. 

Since the total of the angles at the center of any circle 
is 360 degrees and the circumference is 2 nr, where 
7r = 3. 1 41 6 and r = radius, 360 degrees = 2 wr (central 
angles are measured by their arcs). 

?6o° 180 

2 7T 7T 

That is, an arc which equals r, subtends an angle of 
5 7. 3 , or, more accurately, 2 06, 2 65 ". 

Since the total circumference is 2 tt times r, and r 
represents a radian, the circumference contains 27r radians ; 
also the arc subtending an angle has the same ratio to the 
entire circumference that the angle has to 360 degrees. 
Hence the angle will contain the same part of 2 it radians 



Solution of Oblique Triangles. 221 

that it does of 360 degrees, or, what is the same thing, it 
will contain the same part of tt radians that it does of 
180 degrees. 

Find the value of 30 , 45 °, 65 °, 90 , 225 in radians. 

30 = -^or I of 180 
* 180 6 

.\ -\o° = -^ — 7r radians = — radians. 
180 6 

45 - ^5-or lof 180 
180 4 

.*. 45 = ^"5 it radians = — radians. 
180 4 

65 = --5- or H. of l8o o 
180 36 

.*. 65 = —2- 7r radians. 

oo° = -9^_ or J of 180 
180 

.*. 90 = — radians. 
2 

225°= -^-or iof 180 
180 4 

.*. 225°= ^radians, etc. 
4 

Express, 22} , 40 , 135 , 300 , 270 in radians. 

An angle is plainly the same part of 180 that it is 
of 7r radians, hence the process is the exact reverse of 
the above. 

For example , — radians = —of 180 = 6o° 
3 3 

— radians = — of 180 = 72 , etc. 
5 5 



224 Plane Trigonometry. 

1. Express in radians, 130 ; 90 ; 75 ; 225 ; 67^ 

11° I 5 '; 3 I2°; 7 20°; 3 2° 12' 20-. 

2. Express in degrees, — -n- rad. ;— rad. ; .23 ir rad. ; 

3 6 

2.\ it rad.; .25 7r rad. ; f tt rad. 

3. If a circular object subtends an angle of i° at a dis- 
tance of 114.6', what is its diameter? 

4. If a wheel makes 20 revolutions per second, what is 
its angular velocity in radians ? 

5. What is the radius of a circle if an arc of 2100 
miles subtends an angle of 57.3 minutes at the center? 

6. If the difference in latitude between two places on 
the earth (regarded as a sphere) is 7 12', and their 
distance apart is 495.8 miles, what is the diameter of the 
earth ? 

7. At 3 o'clock what is the angle expressed in radians 
between the hands of a watch ? 

8. The moon is 239,000 miles from the earth (approx.), 
and its diameter is 2162 miles. What angles does it sub- 
tend to us ? 



PART V. 

SPHERICAL TRIGONOMETRY. 

Article i. A spherical polygon is a portion of the 
surface of a sphere inclosed by intersecting arcs of 
great circles. 

Hence the sides are measured in degrees, minutes, and 
seconds, instead of linear units. 

Knowing the radius of the sphere of whose surface the 
polygon is a part, the length of its sides can be also easily 
expressed in linear units, for any side will be the same 
part of a circumference (found from the formula, 2 irr) as 
its number of degrees is of 360 . 

It is to be remembered that an arc of a great circle 
bears the same relation to a spherical surface that a 
straight line does to a plane surface. 

Art. 2. By Solid Geometry, the sum of the sides of a 
spherical triangle is always less than 360 ; and the sum 
of its angles is greater than 180 and less than 540 . 

Also the essential theorems relating to plane triangles 
apply equally to spherical triangles. 

Right Spherical Triangles. 

Art. 3. As in Plane Trigonometry, the right triangle 
furnishes the simplest relations between its parts, and 
hence it provides the natural starting-point. Let, then, 
ABC (see Fig. 34) (notation being the same as before) be 
a spherical right triangle, with sides a, b, and c ; C being 
the right angle. To avail ourselves of the known rela- 
tions of Geometry, let O be the center of the sphere, of 



226 



Plane Trigonometry. 



whose surface ABC is part. Join O with A, B, and C- 
Through the vertex A pass a plane _1_ to OB, intersecting 
the face OAB in AD, the face OAC in AE, and the face 
OBC in DE. Then since OB is _L to 4£>E, its plane 




Fig. 34. 

OBC is ± to ADE; and hence AE (a. line in ADE drawn 
through a point of the intersection of these two perpendic- 
ular planes) is _L to OBC, and hence is _L to OC and DE, 
lines of the plane OBC. 

That is, AED, AEO, ADO, and ODE are right angles, 
and ADE is the plane angle of the dihedral whose edge 
is OB, or Z ADE = Z£ (by Geometry). 

The radius of this sphere may clearly be taken as 
unity for simplicity's sake, without in any way affecting 
results. 

Remembering that the central angles AOC, AOB, and 
BOC are measured by their arcs, respectively b, c, and a, the 
plane right triangles, AOE, AOD, DOE, and ADE will 
plainly furnish relations between a, b, c, and B. For 
example : 

By Goniometry (since radius = i), cos AOB = cos c 

= OD, but in A ODE, cos DOE 



cos a 



OE, cos a = cos b cos a, since OE 



OE 
cos AOC - 



or OD 

cos 6. 



cos c = cos ft COS # 



(0 



Spherical Trigonometry. 227 

Formula (1) may be stated thus : In a right spherical 
triangle the cosine of the hypotenuse equals the product 
of the cosines of the two legs. 

It bears the same relation to the spherical right triangle 
that the Pythagorean theorem does to the plane right 
triangle. Again, 

sin c = AD = AE = -^-^- orsin b = sin c sin B (2"). 
sin ADE sin B 

Put this formula into a rule. 

By changing the construction of the figure (drawing 
the perpendicular plane through B), it can be similarly 
proved that, sin a = sin c sin A (2 h ). This formula 
could be inferred by analogy. 

Again, 



tan a 


_ DE _ 
OD 


AD cos B 
OD 


sin c cos 
cos c 


B 


sin c n 

= x cos B 

cos c 




= tan r. 1 


cos B . 








(3°) 
(3*) 


By analogy, 


tan b = tan c cos A 






Again, 














cos B 


DE 
AD 


OE sin a 
AD 


cosb sin 
sin c 


a m 


but from 


(2») 


sin a 


= sin A 


. .-. cos B 


= cos b X 


sin a 




sin c 








sin c 






= cos b 


sin A 








(4°) 
(4 6 ) 


By ar 


lalogy, 


cos A = 


cos a sin 


B 





Art. 4. Thus a variety of combinations may be made 
and each relation proved geometrically. 

By grouping and comparing these various formulae, 
Baron Napier, a famous Scotch mathematician, discovered 
a very simple device for reproducing them. Understand, 
his rules are purely empirical, that is, found by trial, and 



228 Plane Trigonometry. 

are not proofs in any sense ; but since these formulae can 
be proved rigidly, Napier's rules make their reproduction 
easy. 

Ignoring the right angle, and taking the other five parts 
in a circle just in the order they occur, but using the 
complements of the two angles and of the hypotenuse, 
the rules are as follows (see Fig. 35) : 

(1) The sine of any part is equal to the product of the 
tangents of the two parts adjacent to it. 

(2) The sine of any part is equal to the product of the 
cosines of the two parts opposite (not adjacent) to it. 

For example : 

sin (Co. A) = tan (Co. c) tan b. 
Since, sin (Co. A) = cos A and tan (Co. c) = cote, 

cos A = cot c tan b, 
which is formula (3 5 ) already found. 



Co.A 



Co.B 




Fig. 35- 

Again, sin (Co. A) = cos (Co. B) cos a, 

or, cos A = sin B cos a, which is formula (4^). 

Again, sin (Co. c) = cos a cos b, 

or, cos c = cos a cos b, which is (1), etc. 



Spherical Trigonometry. 229 

Art. 5. It is to be observed that if three certain parts 
are to be combined in an equation, one must be chosen, 
to which the others are either both adjacent or both oppo- 
site, in order to use Napier's rules. 

Suppose, for example, A and c are given in a right 
spherical triangle and the other parts are required. The 

case stands thus: >• given f . , 

c j a > required. 

b ) 

First, to find B : since A and c are the known parts and 
B is to be found, an equation between A, c, and B is 
necessary. Of the three, A, c, and B, A will not answer 
for the middle part, for c is adjacent and B opposite; c, 
however, has both A and B adjacent to it, hence by rule 1 : 

sin (Co. c) = tan (Co. A) tan (Co. B) 
or, cos c = cot A cot B 

whence, cot B = — : — = cos c tan A (whence B is found) . 
cot A 

Second, to find a : of the three, A, c, and a, A cannot be 
middle, nor can c, for in neither case do the other two 
occupy the same position relative to it; but a has both A 
and c opposite to it. Hence, by rule 2, 

sin a = cos (Co. A) cos (Co. c) 
or, sin a = sin A sin c (whence a is found). 

By a like procedure, find the formula for b. 

Art. 6. A quadrantal triangle is one having at least 
one side a quadrant (90 ) in length. Its solution can be 
easily reduced to that of a right spherical triangle by using 
its polar. 

Example. Solve the triangle in which a = 90 , B = 
65 °, c = 8o°. Constructing the polar triangle and calling 
corresponding sides and angles a\ b f , c\ A', B', C, we 



230 Plane Trigonometry. 

have by Geometry, a' = 180 — A, b' = 180 — B, c' = 180 
- C, A' - 180 - a, £' = 180 - ft, C" = 180 - c; hence 
4' = 180 - 90 = 90 and A'B'C is a right spherical 
triangle, a' being the hypotenuse. 
To find b. 

.-. sin (Co. B f ) = cos (Co. C) cos ft' or cos B' 
= sin C" cos &' 

whence, cos (180 — 5) = sin (180 — c) cos (180 — B) 
or, — cos b = (sin c) ( — cos B) = — sin c cos i? 

that is, cos b = sin c cos 5, etc. 

EXERCISE I. 

Right Spherical Triangle (C = 90 ). 

b = 69 21' 13" 

4 = 88° 14' 17" 

c - 6i° 23' 27" 

a = no° 17' 24" 

* = 79° 19' 19* 
a = 25 32' 47" 
b = 58 8" 
104 10' is 7 ', 5 = 70 16' 26" 
5 = 112 3 8' 10" 

fl- 99° "' 33" 
B = 78 29' 14" 

12. ^ = 83 44' 22", c = io° 19' 25" 

13. b = 32 47' 18", A = 8o° 30' 20" 

14. a = 29 18' 18", 5 = 142 39' 27" 

15. A = 152 21' 21", 5 = 149 7' 9" 

16. a - 90 , c = 94 20' 37", b - 75 15' 28" 

Isosceles Triangle. 

Art. 7. The spherical isosceles triangle depends upon 
the right spherical triangle for solution in exactly the same 
way that the plane isosceles triangle depends upon the 



1. 


a = 


39° 27' 32", 


2. 


B = 


112° IO' II", 


3. 


b = 


56° 25' 42", 


4. 


A = 


76° 3°' 52", 


5. 


A - 


67° 29' 39", 


6. 


B - 


42° 47' 58". 


7. 


5 = 


98° 45' 46", 


8. 


c = 


90 , C = 10, 


9. 


c = 


163 14' 12", 


10. 


£ = 


102 27' 6", 


11. 


A - 


53° 49' 36*. 



Spherical Trigonometry. 



231 



plane right triangle. An arc of a great circle drawn 
through the vertex perpendicular to the base, divides the 
isosceles triangle into two equal right triangles, which are 
readily solved. 

Art. 8. The analogy in process between plane and 
spherical trigonometry is maintained in the solution of an 
oblique spherical triangle, which is made to depend upon 
the solution of right spherical triangles, by drawing an arc 
of a great circle through one vertex perpendicular to the 
opposite side, thus forming two right triangles, which, 
however, are not equal unless the original triangle is 
isosceles. 

Art. 9. To solve an oblique spherical triangle it is 
necessary to develop certain relations between the parts 
by the use of these right triangles. It is to be observed 
that there is no such simple relation between the angles 




as exists in plane triangles, and hence it is necessary by 
appropriate formulae to find each part by calculation. 

Let ABC (Fig. 36) be a spherical triangle, with sides a, 
b, c. Through C draw the great circle arc CD J_ to AB 
at D. Call CD, h\ AD, m; and DB, n. 

In the right triangle A CD (b being the hypotenuse), 
taking h as a middle part, with A and b ; sin h = cos 
(Co. A) cos (Co. b) = sin A sin b. 



232 Plane Trigonometry. 

Likewise, in CDB ; sin h — sin B sin a. 

.-. smA sin b = sin B sin a, 
or, sin A : sin 2? : : sin a : sin b (i TO ). 

Similarly, sin ^4 : sin C : : sin a : sin c (i w ) 
sin 2? : sin C : : sin b : sin c (i°) . , 



Two angles and 
opposite side, or 
two sides and 
opposite angle. 



Put these formulae into the form of a rule. 

Art. 10. Again, in the right triangle CBD, 
cos a = cos h cos n = cos h cos (c — m), (since n = c — m) 
= cos h cos <; cos m + cos /^ sin c sin m (1) [cos {c — m) 
= cos c cos m + sin c sin m], but cos h cos w = cos b 
(in the right triangle ACD) and, 

7 • cos b 

cos fe sin w = ■ X sin m 

cos w 

[for cos b = cos Ji cos m, .*. cos h = 
cos m\ 

7 sin m 7 ^ 7 cos ^4 

= cos b X ■ = cos b tan w = cos b 

cos m cot b 

= sin £ cos A for cos ^ = cot b tan m, 



cos A t cos 6 . ~ 

tan m — , also = sin b 

cot b cot b 



Substituting these values for cos h cos m and cos h 
sin m in (1), 

cos a = cos b cos c + sin b sin c cos A . . . (2"') 

By similar process or by analogy, 
cos b = cos a cos c + sin a sin c cos J5 . (2") 

cos c = cos a cos & + sin a sin 6 cos C . . . (2 ) 

Napier's rules are applied in every case above. 



Spherical Trigoiiometry . 233 

Art. 11. From (2™), solving the equation for cos A, 

cos A = cos a ~ cos b cos c (*) 

sin b sin c 

Hence, 

A _ cos a — cos b cos £ (subtracting both 

I — COS A — I — ... 

sin b sin c sides from 1) 
sin b sin c + cos & cos c — cos a _ cos (jb — c)— cos a 
sin b sin c sin 6 sin c 

cos (6 — c) — cos a _ — 2 sin % (6 - c + a) sin % (b — c — a) 
sin 6 sin c sin & sin c 

[By formula 18, Plane Trigonometry, calling P = (b — c)~\ 
and Q = a, 

- 2 sul £ (b — c + a) sin -| (q — b + c) 
sin 5 sin c 

[for — sin x = sin ( — x) , hence, — sin ^ (6 — c — a)~\ 
= sin ^ [ - (b-c — a)] = sin ^ (q — 6 + c) . 

Let 5 = i (a + b + c) 

then, s — a = % (b - a + c) 

s — b = h {a — b + c) 

s — c = % (a + b — c). 

Substituting these values above, 

. 2 sin \ (b — c 4- a) sin h (a — b + c) 

I — COS A = — ; L — L 

sin b sin c 
_ 2 sin \ (s - c) sin h (s — b) 
sin b sin c 

but, 1 — cos A = 2 sin 2 ^ ^ (by Goniometry), 

- 2 1 a 2 sm £ (s - c ) sm + (5 - 6) 

,\ 2 sin 2 \ A = J v . '- 4-j ' 

sin 6 sin c 

or, sin' j A = sin * ( * ~ f ) sin * < f ~ ») . , (3") 

sin 6 sin £ 



234 Plane Trigonometry. 

By a similar process with (2 n ) and (2 ) or by analogy, 

. , , „ sin \ (s — a) sin I (s — c) , ■ „ 

sm 2 \ B = a - i ^ — L . . (3*1 

sin a sin c 

sin 2 \ C = sin i fr - <0 sin i fr ~ &) , # ( 3 <a 
sin a sin 6 

By adding 1 to each side of equation (x) the value of 
1 + cos A = cos 2 J A can be easily found to be : 



cos 



1 A = sin * sin ~ g ) 8 . . . U m ) 

sin 6 sin c 

Likewise, cos* |J - sin > sin( f ~ 6) .... (4") 



sm a sin c 



, 21^- sin s sin (s — c) , ON 

and cos 2 J C = : 7 L .... (4 ) 

sin b sin a 



Dividing (3-) by ( 4 W ) j ( 3 n ) by ( 4 n ) ; (3 ) by (4 ), 

sin 2 J A = tan2 x A= sin(j-c)sin(j-6) ^ . 
cos 2 \A sin 5- sin (5 - a) 

tan' j g = sin ( * ~ a ) sin (s - c > . . ( S ") 

sin 5 sin (5 — £>) 

tan' tC' Si " (s " a) Si " ( ' " &) • (5°) 

sin s sin (5 — c) 

Art. 12. (5™), (5 n ), (5 ) have the least common 
multiple, 

sin (s - a) sin Q - b) sin (5 - c) _ , . 



say 
sin s 



-* 



sin (5 - g) sin (■> - b) sin (5 - c) 
sin 5 



Spherical Trigonometry. 235 



If r 2 be divided successively by (5" 1 ), (5 n ), and (5 ) 
and the roots of the quotients extracted, the results are : 



tan J A = i- . . . (6 m ) 

sin (s — a) 

tan i B = ■£■ . . . (6") 



sin (s — 


a) 


r 




sin (s - 


b) 


r 





Three sides. 



taniC = — y . . . (6°) 

sin (s - c) 

Attention is called to the analogy between these re- 
sults and the corresponding formulae under Plane Trigo- 
nometry. They will be found to have exactly similar 
application, and the use of r is as before a great simpli- 
fication of the labor in solution of triangles. 

Art. 13. Reverting to formula (2" 1 ), 

cos a = cos b cos c + sin b sin c cos A t 

and substituting the values of a, b, c, and A in terms of 
the sides and angles of the polar triangle, (x) becomes 

cos (180 - A') = cos (180 - B') cos (180 - C) 

+ sin (180 - B') sin (180 - C") cos (180 -a') 
or, - cos A ' = ( - cos B') ( - cos C) + (sin B') (sin C") 

(-cos a') 
or, cos A' = - cos B' cos C' + sin B' sin C' cos a'. 

It is clear that the accents have no significance except 
to distinguish the parts of one triangle from the corre- 
sponding parts of its polar. Since a relation has been 
found between the parts of this single triangle among 
themselves, and since this triangle, although it happens 
to be polar to a certain other triangle, is not in any sense 
a special kind of triangle, the above result is perfectly 
general, and the accents may be dropped ; hence, 

cos A = -cosB cos C + sin B sin C cos a. (j m ) 



236 

Likewise, 



Plane Trigonometry. 



cos B = - cos A cos C + sin ^4 sin C cos & . (7") 
cos C= -cos ^4 cos B + sin ^4 sin B cose . (7 ) 

Art. 14. By treating (f l ), (y n ), and (7 ) exactly as 
we did (2™) , the following formulae arise : 



sin'jq- ~ cos -S cos (5«- ^) 
sin i? sin C 



(8-) 



sin 



2 j , _ - cos 5 cos (S — B) 
sin A sin C 
sin2 c= - cos 5 cos (5 - C) _ 
sin A sin ,6 



cos 



21 
2 



cos' 



COS 



cos (5 - B) cos (S - C) 

sin B sin C 

15= cos (5 - ,4) cos QS - C) 

sin y4 sin C 

2 1 = cos (5 - A ) cos (6 1 - B) 

sin ^4 sin B 



— cos 5 cos (S — A) 
cos (5 - B) cos (5 - C) 

— cos 5 cos (S — B) 
cos (5 - A) cos (5 - C) 

— cos 5 co s (£ — C) 
cos (5 - A) cos (5 - B) 



tan 2 J a = 
tan 2 £ b = 
tan 2 £c = 



[where S = j~| 
• (8") 



• (8°) 

• (9 m ) 

• (9 n ) 

• (9°) 

• (io n ) 

• (io°) 



The G. C. D. of (io w ), (io w ), and (io°) is found to be 

— cos S D2 

= R 2 , say. 

cos {S - A) cos {S - B) cos (5 - C) J 

:. tan \a = R cos (5 - 4) . . (n w ) 

tan £ 6 = R cos (5 - B) . . (n n ) J- Three angles. 
tan£c = 2? cos (S - C) . . (ii°) 



Spherical Trigonometry. 237 

Note. — It is to be observed that A + B + C = 2 S is 
always greater than 180 and less than 540, by Geometry, 
and hence S is always greater than 90 and less than 
270 ; and hence cosS is always negative, by Goniometry. 
Therefore, — cos S must be always positive ; so that the 
values of the radicals in this last article are never imagi- 
nary in a real triangle. 

Art. 15. Dividing (5"*) by (5") we get 

tan 2 \ A _ sin 2 (s - b) 
tan 2 \ B sin 2 {s — a) 

tan I A sin (s - b) 

or, 2 — = ) ' 

tan \ B sin {s — a) 

or, tan \ A : tan'£ B : : sin (s - b) : sin (5 - a). 

By composition and division, 

tan \ A + tan \ B : tan \ A — "tan \ B : : sin (s — b) 

+ sin (5 - a) : sin (j - b) -sin - a), 



whence, 

tan \ A - tan \ B _ sin ($-&)- s in (. v - a) 

tan % A + tan ^ 2? sin (s — b) + sin (.9 - a) 

But 

sin J ^ _ sin ^ i? 

tan ^ ^ - tan \ B _ cos % A cos | 2 ? 

tan $ A + tan J £ sin ^ ^4 sin ^ B 

rr>Q J- /I r^n<; X R 



(P) 



s 1 - 

cos ^ ^4 cos ^ B 
sin £ ^ cos \ B - cos J ^4 sin J Z? 
sin £ ^4 cos I B + cos £ 4 sin ^ 5 
sin (fr 4-^,5) sinh(A-B) , n , 
sin (f 



1 £ ^ cos £ Z5 - cos % A sm % £J 
1 £ A cos I 5 + cos £ A sin £ £ 
>(»4-}*) = sin l(A-B) (p v 
1 (hA + iB) sinh(A + B) 



2 3 8 



Plane Trigonometry. 



Again, 

sin (s — b) — sin (s — a) _ 2 cos % (2 s —a — b) sin i (a — b) 

sin (5 - 6) + sin (s - a) 2 sin 1 (2 s - a - b) cos i (a-b) 



["Let P =.(s - b) and Q =1 

[(5 - a) in (15) and (16) j 



sin Jc cosi (a-b) J 2j 

["Since 2s=a + b + c, .-."] 
\_2s~a — b = c ) 

.-. substituting (P t ) and (P 2 ) in (P) 

_ — 2_i — - — 1 = co t 1 c tan \ (a - b) ; or tan A (a - b) 
sin £ (i4 + B) 

smh (A - B) . , , „,. 

= — — ^— ; ^-tan *c ... (i2 m ) 

sin 1 (A + 5) . ; 

[Two angles and included! 
side. j 

By the same process, using (5") with (5 ) and (5™) with 
(5°) ^ we g et > 
tan \ {b - c) = Sm ; ~ ( tan 1 a Two angles (i2 n ) 
S . m >7_y hand included 

tan 1 (a - c ) = ^-^ ^tan J 5 side. (12 ) 

sin (4 + C) J 



Art. 16. Using the polar triangle, and substituting in 
(i2 m ), (i2 w ), and (12 ) the values of their parts in terms of 
the supplementary parts of the polar as was done in Art. 
13, arise the corresponding formulae : 



sin ^ (a + 0) 

tan J (B - C) = Sin t^~^ c °H^ 
sm J (5 + c) 

tan J W - C) = S ? n t^~^ cot^^ 
sin ^ (a + c) 



Two sides and 
included angle 



(i3 m ) 
(i3") 
(13°) 



Spherical Trigonometry. 



239 



Art. 17. Multiplying together (5™) and (5"), 

tan 2 \ A tan 2 i£= sm2 ^ ~ ^ 
2 2 sin 2 5 

tan \ A tan \ B sin (^ — c) 

or, ^ ^ — • = ^ '-. 

1 sin s 

As before, taking this proportion by composition and 

division : 

1 + tan \ A tan \ B _ sin s + sin (s — c) 



But 



1 — tan ^ A tan ^ B sin s — sin (s — c) 

, sin * ^4 sin -A- B 

1 + * ^— 

1 + tan % A t an \ B _ cos ^ ^4 cos \ B 

1 — tan \ A tan ^ 5 sin \ A sin ^ B 



(y) 



cos \ A cos ^ B 
_ cos -| .4 cos \ B + sin \ A sin ^ 2? _ cos \ (A — B) 
cos \ A cos \ B — sin \ A sin J B cos ^ (^4 + B) 
Also, 

sin .? + sin (s — c) _ 2 sin \ (2 s — c) cos ^ c 
sin j — sin (s — c) 2 cos i (2 s — c) sin ^ £ 

[substituting s for P, and"! 
- c) for Q in (15) and(i6) J 
= tan % {a + b) cot ^ c [since 25 = alHc; 25 
— c = a + b] 
,\ (y) becomes, substituting these values for its members, 
cos j (A — B) 
cos J (A + B) 
or, 

tan £ (a + b) 

Likewise, 

tan l(b + c)= C0S i[ B n ~ S tatU 

and 

tani (a+r) = - HU ~ C) tanP 



tan 4 (a + 6) cot | <;, 



cos 1 (4 - B) 
cos,V (,4+£) 

cos^(^-C) 
cosi(£ + Q 

cos \ (A-C) 
cos h (A+C) 



tan A r 



(12*) 

Two angles 
and included ( I2 ») 
side. 



(12=) 



240 



Plane Trigonometry. 



By using polar triangle in application to (i2 x ), (i2 ?/ ), 
(12 s ), we get, 



tan±{A+B)= COSi( ?- b) cot±C 
cos ^ (a + b) 

tani(3 + C) = cos f V-*) cot I A 

cos -| (b + c) 

^ v ; cos£(<z + c) 2 



(13*) 

Two sides and . 
included angle ^3 >) 

(i3 2 ) 



Art. i 8. By using formulae (i m ). (i n ), (i°), when 2 
sides and an opposite angle, or 2 angles and an opposite 
side are given ; formulae (6 m ) , (6 n ) , (6°) , when 3 sides are 
given ; formulae (n m ), (n n ), (n°), when 3 angles are 
given; formulae (i2 x ), (i2 y ), (12 2 ), when 2 angles and the 
included side are given; formulae (13*), (13*), (13*), when 
2 sides and the included angle are given, any spherical 
triangle may be completely solved. 

Art. 19. Example. Given A = 135 21' 21"; a = 
117 10' 18"; b = 78 23' 40". 

To find B. sin A : sin B : : sin a : sin b 

sin A sin 6 



sin B 



sin a 



log sin B = log sin A + log sin 5 + colog sin a 

log sin A = log sin (180 — A) 

= log sin 44 38' 39" = 9.846771 - 10 

log sin 78 23' 40" = 9.991029 — 10 

colog sin a = colog sin (180 — a) 

= colog sin 6 2 ° — 4 9 — 42"= 0.050785 

log sin B = 19.888585 - 20 

= 9.888585 - 10 



B = 50 41' 21". 



Spherical Trigonometry. 2 4 J 

To find c. Formula (i2 m ) contains the known parts a, 
b, A, B and the unknown c, with no others ; hence by solv- 
ing it for tan ^ c, the value of c may be found. 

In (i2 m ) then, tan J c = sm I U + B) t i (a - 6) 
V ; 2 sin i (4 - 5) 

4 = 135 21' 21" a = 117 io' 18" 

£ = 50 4 i' 2i ff & = 78 23' 40" 

i + B = 186 2' 42" a - b = 3S 46' 38" 

4 _ B = 84 40' 00" i (a - b) = 19 23' 19" 
i (A + B) = 93° i' 21" 
£ (4 - JS) - 42 20' 

log tan \c = log sin J (.1 + B) + colog sin % (A - B) 
+ log tan h {a — b) 

log sin \ {A + B) = log sin [180 - \ (A + B)] = log 
sin 86° 58' 39" = 9.999396 - 10 

colog sin ^ (.4 - B) = colog sin 42 ° 20' 

= 0.171699 

log tan £ (a — b) 

= log tan 19 23' 19" = 9.546459 - 10 

log tan \c = 9.717554 - 10 
ic =27° 33' 30" 
< = 55° 7' 



To find C, use formula (13'") in the same way. 

Example. Given A = uo° 36' 24"; £ = 122 8' 
42"; C = 140 20' 18". 

Here the three angles are given to find three sides, hence 
formula; (11'"), (n n ), (n°) apply- 

2 log R = log ( - cos 5) + colog cos (S - A) 
+ colog cos (5 - B) + colog cos ($ - C) . 



242 Plane Trigonometry. 





A = no 3 6' 24" 






5 = 122° 8' 4 2 r/ 






C = 140 20' 18" 






2 S = 373 - 5 ~ 2 4 






6" = 186 — 32 — 42 






S ~ A = 75 - 56 - 


- 18 




6* — B = 64 — 24 - 


■ 00 




S - C = 46 - 12 - 


• 24 


log ( - cos S) 


= log - [ - cos (180 - 


S)] 




= log cos 6° 32' 42" = 


9.997180 


colog cos (S 


-A) 






= colog cos 75 56' 18" = 


0.614152 


colog cos (S - 


- B) = colog cos 64 24' = 


0-36443° 


colog cos (S - 


- C) 






= Colog COS 46° I2 r 24" = 


°.i5975i 




2 log £ = 


1 - 1 355 1 3 




log 12 = 


•5°77565 


log tan \ a = 


= log R + log cos (6" - 


-4) 


log tan J & = 


= log R + log cos (5 - 


B) 


log tan \ c = 


= log R + log cos (5 - 


C) 




log R = 


•5677565 




log cos (5 - 4) = 


9.385848 



— 10 



log tan 1 a = 9.9536045 - 

ia = 4i°56 / 43" 
a = 8 3 ° 53' 26" 
log 12 = .5677565 
log cos (S - B) = 9-63557o 

log tan \ b = 10.2033265 — 

*&-.57° 56' 51" 

ft = 115 53' 42" 
log R = .5677565 
l og cos (S — C) = 9.8 40349 

log tan \ c = 10.40810155 - 
I c = 68° 39' 26" 
c- i37°i9' 5 2 " 



Spherical Trigonometry. 243 

EXERCISE II. 

Find unknown parts of following triangles : 

1. a = 57°56 , 42 / ' b = 137° 22' 18" C = 94° 47' 12" 

2. ^ = 131° 17' 24- C= 94° 48' 24" a- 57° 56' 36" 

3. ,4= 68° 34' £=130° 48' 24" C = 94° i'36* 

4. a =149° 24' 24" b = 129 48' 24" c= 67° 19' 12" 

5. <; = 88° 12' 20" 6=124° 8' 17" C= 50 2' 1" 

6. ,4= 76° 13' 42" b= 96° 49' 6" c= 83° 18' 25" 
1. a = 48° 48' 48" B= 139° 20' 30" c- 84°3 9 , 2 9 // 

8. 4 = 65° 41' 16" 5=io 9 ° 33' 22" c= 78° 42' 36" 

9. j B = iii°44 , 46 // &=io2°37'i4* C = 8 9 °27'i 5 " 

10. a . 8 3 °4o / 4o' 7 B= 68° 18' 17" C= 4 9 °ii , io // 

11. b= 26° 5 6' 4 8 // 5= 39°io , 45 // C= i45°35'36" 

12. a =100° 47' 9" 5= 99°36 / i3 ,/ c - 87° 49 , 27 ,/ 

13. ,4 = i27 32'2 5 " J B=ii2 57' 4 2" C- 75°55 , 45 // 

14. a = 6S°38 / 48" 6 = 73°42 , 37 // c = 58° 17' 16 J" 

15. C=ii3°io' 7" 5= 98°43 / i 4 ' / c= 7 i°2i / 8" 

16. 0- 39° 7' 7" 6= 77°33'"" C- 82°2 3 ' 5 2" 

17. 5 = io 9 22 / ii" 6= 6o°47 , i 7 // C=io2°37 , i 9 ' / 

18. .4 = 133° 6' 4" 5= 9i°48 , 2 4 // C= 78°4 3 ' 5 8" 

19. b= 44° 33' 20" £ = 86° 25' 18" c- 22° 16' 40" 

20. ,1= 97° 27' 32.4"/;= 62° 14' 17^= 59° 52' 4" 

21. (7 = 49°57 , 57" 6= 5 l0 42'37" c= 8 2 ° 4 2'i8" 

22. .4 = 112° 46' 33" £=109° 27' 23" C= 98° 7' 36" 

23. A= 56° 56' 5 6 // 6= 79° 28' 43" c = 62° 30' 21" 

24. B = i2i°i 9 ' 39 " a= 8i°57'i6" C= 85°47' 3 2" 



PART VI. 



APPLICATION OF SPHERICAL TRIGONOMETRY. 

Article 20. Most frequent application of Spherical 
Trigonometry is made in Practical Astronomy and Navi- 
gation. 

In these sciences, the Spherical Triangle takes a specific 
form, known as the Astronomical Triangle, the earth's 
surface being regarded as spherical, and hence the 
meridians as great circles. 

In the accompanying figure, 37, ABC is the horizon; 




FGH is the equator projected on the sky, called the 
equatorial ; PZA is the meridian. 

The observer is supposed to be at the center, O, the 
point Z being the zenith, and P the north pole. S, being 
any celestial object, SM is called its declination, (d) ; SN 
its altitude, (h) ; PL the latitude (I) of the observer at O. 

Hence in the Spherical Triangle ZSP, ZS = 90 — h ; 
SP = 90 - d\ ZP = 90 - /; also the angle AZN be- 

244 



Application of Spherical Trigonometry. 245 

tween the great circle (vertical circle) through S and Z, 
and PZA the meridian, reckoned from the south point 
through the west point, is called the azimuth, (a) of S. 

The angle ZPS, between the great circle (hour circle) 
PSM through 5 and P, and the meridian, PZA, is called 
the hour angle, (/) of S. 

Hence in the triangle PZS = 180 - AZN (in the 
figure) and AZN is found by subtracting the azimuth 
from 360, when it exceeds 180 . 

Art. 21. Example. What time does the sun set in 
St. Petersburg, lat. 59 56', on the longest day of the 
year? 

On the longest day of the year the sun is farthest north, 
and its declination on that day is always 23 27', the 
angle its apparent path makes with the equinoctial. Also 
at setting it is on the western horizon, hence its latitude 
is zero. .-. in the triangle ZSP, ZP = 90 - / = 90 

- (59° 56') - 30 4'; SP = 90 - d - 90° - (*3° 27') 
= 66° 33' ; and SZ = 90 - h = 90 . 
We have then a triangle with three sides given, which is 




Fig. 38. 



solvable by the method explained previously, or since 
it is also a quadrantal triangle, we can use its polar which 
will be a right triangle. 

Then in the figure (38), drawing the polar x y z, since 



246 Plane Trigonometry. 

the hour angle t is required, we must find side /'; 
t' = 180 - t. 

y = 180 - 90 = 90 ; z = 180 - (90 - I) = 90 + / 
x = 180 - (90 - d) = 90 + d, and t' is hypotenuse. 

By Napier's rules, t f being a middle part to x and z the 
other known parts, sin (co. /') = tan (co. x) tan (co. z) 
or cos t' = cot x cot z. Substituting values above, cos 
(180 — /) = cot (90 + d) cot (90 + /), 

whence, — cos / = ( — tan d) ( — tan /) , or cos t 
= — tan d tan /. 

That is, cost = - tan (23 27') tan (59 56') 

tan 23 27' = 9.63726 — 10 
tan 59 56' = 10.23739 - 10 
— cos t = 9.87465 — 10 
/ = 180 - (48 31' 44") = 131 28' 16" 
or in time, / = (13 1° 28' 16") -- 15 

= 8 hr. — 45 min. — 53 sec. 

That is, the sun sets about 8.46 o'clock P.M. 

Again : On a given day the sun's declination is 
18 35 r N. At 3 o'clock P.M. its altitude is 48 22'. 
What is the latitude of the place ? 

In the triangle ZPS, we have here, ZPS (t) = 45 ° [3 
X15;] SP = 9 o°-d = 9 o°- [i8°35']= 71 25'; and 
ZS = 90 - h =9o -[48 22 r ] = 41 38', to find ZP = 90 
— /. That is, we have two sides and one angle given, from 
which the third side ZP is readily found. 

Art. 22. Since the longitude of a place is the same as 
the difference between its local time and Greenwich time, 
if Greenwich time is known at any observation, the hour 
angle as calculated above will give local time, and hence 
the longitude is easily found. 

Every ship carries chronometers with Greenwich time, 
and therefore this method gives its longitude readily. 



Application of Spherical Trigonometry. 247 



Art. 23. There is another class of problems whose 
solution is much simplified by the use of Spherical 
Trigonometry. For example, let it be required to find the 
angle between the lateral faces of a regular octagonal 
pyramid, whose edges meet at an angle of 18 at the 
vertex. 

In the pyramid ABCDEFGH - K to find the angle be- 
tween the faces, say between ABK and CBK. Take B 
as the center of a sphere of 
any convenient radius ; the 
surface of this sphere will in- 
tersect the three faces ABK, 
CBK, and ABCDEFGH in 
the sides of a spherical tri- 
angle, which will be isosceles, 
because the pyramid is reg- 
ular. 

Call this triangle MNP (as 
represented in Fig. 39), the 
sides being m, n and p, ac- 
cording to our usual designa- 
tion. 

By geometry, Z. ABC = 
arc p = i [2 right angles 
X (8 - 2)] = I right angles = 135 . Also, since the 
pyramid is regular, KBC (or KB A) is isosceles. Hence, 
since BKC (or BKA)= 18 , KBC (or KBA) = i (180 
- 18 ) = 8i°. That is, arcs m and n = 8i°. 

By dropping a perpendicular arc from P to MN, say 
at R, the isosceles triangle is divided into two equal right 
triangles, wherein m (or n) = 8i° and RN (or RM) = ^ 
(135 ) = 67 30'. Whence | Z. P (the required angle) is 
easily found. 

Again : Through the foot of a rod making an angle m 
with a plane, a straight line is drawn making an angle 
n, with the projection of the rod on the plane. 




248 



Plane Trigonometry. 



What angle does the rod make with this line ? Let 
MN be the plane, and OA the rod, OC its projection, and 
OB the line in the plane. With O as a center describe 
a sphere with any convenient radius. It will intersect 
the planes of the three lines in the right spherical tri- 




Fig. 40. 

angle MNP, whose sides will be m, n, and p, of which m 
and n are known. With m and n known, it will be easy 
to find p, which equals the required angle A OB. Why 
is the spherical triangle a right one ? It will be observed 
that while these problems can be solved by Plane Trig- 
onometry methods, the solutions are greatly simplified by 
the application of the spherical. 



Area of Spherical Triangle. 

Art. 24. By Solid Geometry the area of a spherical 

triangle is given by the formula : , E being the 

180 

spherical excess [180 — (A + B + C)] in the triangle, 

expressed in spherical degrees, and R is the radius of 

the sphere. 

To use this formula it is clearly necessary to know the 

three angles. 



Application of Spherical Trigonometry. 249 

However, the value of E can be found from the three 
sides, by the formula : tan 2 \ E = tan \ S tan \ (S — a) 
tan i (S — b) tan \ (S — c) ; wherein a, b, and c are the 
three sides and S = \ {a + b + c) . 

Example . Find the area of the spherical triangle, 
whose sides are : a = 69 15' 6", b = 120 42' 47", c = 
159 18' $$", on a sphere whose radius is 7918 miles. 

a - 69 15' 6" 

b = 120 42' 47" 

c - i59° i8 / 33" 





a + b + c=2S-- 


= 349° 16' 26" 


s - 

s - 

s - 


S= 174° 38' 13" 
- a = 105 2 3 r 7" 

-*- 53° 55' 26" 

-*= is°i 9 ' 40" 


J (5 -a) - 5 a° 41' 33 i" 

J (5 - 6) - 26 57' 43" 
*<£-*)- 7° 39' 50" 




tan %S 
tan J (5 - 0) 
tan i(S - b) 
tan J (5 - c) 


= 11.32942 - 10 
= 10. 1 1805 — 10 
= 9.70644 — 10 
= 9- I2 893 - 10 



tan 2 J E = 20.28284 - 20 
tan \ E = 10.14142 — 10 
\E= 54 10' 5" 
E = 2i6° 4 o' 20" 



may be expressed , if £ be reduced to sec- 

180 648000 

onds (since 180 must also be multiplied by 3600). 



lo S , Q = 4.68557 - 10 
648000 

log E" = 5.89210 

log7 9 i8 2 = 7-79724 

log area = 8.37491 

area = 237088889 sq. miles. 



2 5° Plane Trigonometry. 

EXERCISE III. 

Applications of Spherical Trigonometry. 

1. A ship's captain observes the sun's altitude to be 
14 1 8' at 6 o'clock A.M. The almanac gives its declin- 
ation as 1 8° 36' N. What is the ship's latitude ? 

2. If a ship in latitude 50 13' finds the sun's altitude 
to be 1 6° 20' at 9 o'clock A.M., Greenwich time, the 
sun's declination being 21 6', what is its longitude? 

3. At what time will the sun rise at Melbourne, lat. 37 
49' S. on the longest day in the southern hemisphere, 
sun's declination being 23 27' S ? 

4. What angle does the shadow on a sun-dial plate 
make with the gnomon at 3 P.M. in latitude 40 37' ? 

5. Find the latitude of the place at which the sun sets 
at 9.30 P.M. on the longest day. 

6. In what latitude will the sun rise exactly in the 
northeast point on the longest day ? 

7. The moon's path makes an angle of 5 8' with the 
ecliptic, in which the axis of the earth's shadow lies. A 
section of this shadow is circular in form with its center 
on the ecliptic. If the radius of the moon is 15 45", 
how far must the moon be from the intersection of its 
path with the ecliptic, that it may just touch the shadow, 
that is, begin an eclipse ? 



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